Mole Concept and Stoichiometry

From equation:

$$\begin{matrix} 2\text{CO} & + & \text{O}_2 & \longrightarrow & 2 \text{CO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$$

[By Gay Lussac's law]

2 V of CO requires = 1V of O₂

∴ 100 litres of CO requires = $\frac{1}{2}$ x 100 = 50 litres.

Hence, required volume of oxygen is 50 litres.

$$\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$$

2 Vol. of hydrogen reacts with 1 Vol. of oxygen

∴ 200 cm³ of hydrogen reacts with = $\frac{1}{2}$ x 200 = 100 cm³ of oxygen.

Hence, unreacted oxygen is 150 - 100 = 50cm³

This experiment supports Gay-Lussac's law of combining volumes.

Since the unchanged oxygen is 58 cc so, used oxygen 106 - 58 = 48cc

According to Gay-Lussac's law, the volumes of gases reacting should be in a simple ratio.

$$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & 2\text{CO}_2 + \text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} \\ 24 \text{ cc} & : & 48 \text{ cc} \end{matrix}$$

Hence, methane and oxygen are in the ratio 1:2 .

$$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$$

[By Gay Lussac's law]

2 Vol. of C₂H₂ requires 5 Vol. of oxygen

∴ 400 ml C₂H₂ will require $\frac{5}{2}$ x 400

= 1000 ml of Oxygen

Hence, required volume of oxygen = 1000 ml

Similarly,

2 Vol. of C₂H₂ produces 4 Vol. of Carbon dioxide

∴ 400 ml of C₂H₂ produces $\frac{4}{2}$ x 400

= 800 ml of Carbon dioxide

Hence, carbon dioxide produced = 800 ml

$$\begin{matrix} \text{H}_2\text{S} & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} & + & \text{S} \\ 1\text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & : & 1\text{ vol.} \\ \end{matrix}$$

(i) At STP,

1 mole gas occupies = 22.4 L.

1 mole H₂S gas produces = 2 moles HCl gas,

∴ 22.4 L H₂S gas produces

= 22.4 × 2

= 44.8 L HCl gas.

Hence, 112 cm³ H₂S gas will produce

= 112 × 2

= 224 cm³ HCl gas.

Hence, 224 cm³ HCl gas is produced.

(ii) 1 mole H₂S gas consumes = 1 mole Cl₂ gas.

Hence, 22.4 L H₂S gas consumes = 22.4 L Cl₂ gas at STP.

∴ 112 cm³ H₂S gas consumes = 112 cm³ Cl₂ gas.

Volume of Cl₂ gas left unreacted = Total volume of Cl₂ gas - Volume of Cl₂ gas consumed

= 120 cm³ - 112 cm³

= 8 cm³

Hence, the composition of the resulting mixture is 224 cm³ HCl gas + 8 cm³ Cl₂ gas.

$$\begin{matrix} 2\text{C}_2\text{H}_6 & + & 7\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 6\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$$

[By Gay Lussac's law]

2 Vol. of C₂H₆ requires 7 Vol. of oxygen

∴ 300 cc C₂H₆ will require $\frac{7}{2}$ x 300

= 1050 cc of Oxygen

Hence, unused oxygen = 1250 - 1050 = 200 cc

Similarly,

2 Vol. of C₂H₆ produces 4 Vol. of carbon dioxide

∴ 300 cc C₂H₆ produces $\frac{4}{2}$ x 300

= 600 cc of Carbon dioxide

Hence, carbon dioxide produced = 600 cc.

$$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow & 2\text{H}_2 & + &\text{O} \\ 2\text{ Vol.} & & 2\text{ Vol.}& & 1\text{ Vol.} \end{matrix}$$

2 Vol. of water gives 2 Vol. of H₂ and 1 Vol. of O₂

∴ If 2500 cm³ of H₂ is produced, volume of O₂ produced = $\frac{2500}{2}$ = 1250 cm³

(b) V₁ = 2500 cm³

P₁ = 1 atm = 760 mm

T₁ = T

T₂ = T

P₂ = [760 x 2 $\frac{1}{2}$] + [760] = 760 [$\frac{5}{2}$ + 1] = 760 x $\frac{7}{2}$ = 2660 mm

V₂ = ?

Using formula:

$\frac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\frac{\text{P}_2\text{V}_2}{\text{T}_2}$

$\frac{760 \times 2500}{\text{T}}$ = $2660 \times \frac{\text{V}_2}{\text{T}}$

V₂ = $\frac{760 \times 2500 }{2660}$ = $\frac{5000}{7}$

(c) V₁ = $\frac{5000}{7}$ = 714.29 cm³

P₁ = P₂ = P

T₁ = T

V₂ = 2500 cm³

T₂ = ?

Using formula:

$\frac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\frac{\text{P}_2\text{V}_2}{\text{T}_2}$

$\frac{\text{P} \times 714.29}{\text{T}}$ = $\frac{\text{P} \times 2500}{\text{T}_2}$

T₂ = $\frac{2500 }{714.29}$ x T

T₂ = 3.5 x T

Hence, T₂ = 3.5 times T or temperature should be increased by 3.5 times

$$\begin{matrix} \text{CH}_4 & + & 2\text{Cl}_2 & \longrightarrow & \text{CH}_2\text{Cl}_2 & + & 2\text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1 \text{ vol.} & : & 2 \text{ vol.}\\ \end{matrix}$$

volume of HCl gas formed = ?

[By Gay Lussac's law]

1 Vol of methane produces = 2 Vol. HCl

∴ 40 ml of methane produces = 80 ml HCl

volume of chlorine gas required = ?

For 1 Vol of methane = 2V of Cl₂ required

∴ for 40 ml of methane = 40 x 2 = 80 ml of Cl₂ is required.

Hence, volume of HCl gas formed = 80 ml and chlorine gas required = 80 ml

Given, oxygen is 1⁄5^th of air = $\frac{1}{5}$ of 500 = 100 cm³

$$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3 \text{ vol.} & \\ \end{matrix}$$

[By Gay Lussac's law]

5 Vol. of O₂ requires 1 Vol. of propane

∴ 100 cm³ of O₂ will require = $\frac{1}{5}$ x 100 = 20 cm³

Hence, propane burnt = 20 cm³ or 20 cc

$$\begin{matrix} 2\text{NO} & + & \text{O}_2 & \longrightarrow & 2\text{NO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$$

[By Gay Lussac's law]

1 Vol. of O₂ reacts with = 2V of NO

200 cm³ oxygen will react with
= 200 × 2
= 400 cm³ of NO

∴ remaining NO is 450 - 400 = 50 cm³

NO₂ = ?

1 Vol. of O₂ produces 2 Vol. of NO₂

∴ 200 cm³ of oxygen produces = $\frac{2}{1}$ x 200 = 400cm³

Hence, NO₂ produced = 400 cm³ and unused nitrogen monoxide (NO) is 50 cm³, so total mixture = 400 + 50 = 450 cm³

$$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$$

[By Gay Lussac's law]

1 Vol. of chlorine reacts with = 1 Vol. of hydrogen

∴ 4 litres of chlorine will react with only 4 litres of hydrogen,

hence, 6 - 4 = 2 litres of hydrogen will remain unreacted.

Since, vol. of HCl gas formed is twice that of chlorine used,

∴ vol.of HCl formed will be 4 x 2 = 8 litres However HCl dissolves in water.

Hence, 2 litres of hydrogen is the residual gas, as HCl formed dissolves in water.

$$\begin{matrix} 4\text{NH}_3 & + & 5\text{O}_2 & \longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} \\ \end{matrix}$$

[By Gay Lussac's law]

9 litres of reactants produces = 4 litres of NO

So, 27 litres of reactants will produces

$$\begin{aligned}=\frac{4}{9} \times 27 \\[0.5em] = 12 \text{ litres}\end{aligned}$$

Hence, volume of nitrogen monoxide produced = 12 litres

According to Gay lussac's law,

$$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$$

As, 4 cm³ of hydrogen was left behind, hence, 36 - 4 = 32 cm³ of mixture of hydrogen and chlorine exploded.

As, 1 Vol. of hydrogen requires 1 Vol. of oxygen
∴ 16 cm³ hydrogen requires 16 cm³ of oxygen

∴ Mixture is 20 cm³ (i.e., 16 + 4) of hydrogen and 16 cm³ of chlorine.

$$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}$$

[By Gay Lussac's law]

1 Vol. CH₄ requires 2 Vol. of O₂

∴ 10 cm³ CH₄ will require 2 x 10

= 20 cm³ of O₂

Given, air contains 20% O₂ by volume.

Let $x$ volume of air contain 20 cm³ of O₂

$$\begin{aligned}\Rightarrow \frac{20}{100} \times x = 20 \\[1em] \Rightarrow x = \frac{100}{20} \times 20 \\[1em] \Rightarrow x = 100 \text{ cm}^3\end{aligned}$$

∴ 20 cm³ O₂ is present in 100 cm³ of air.

Similarly, 2 Vol C₂H₂ requires 5 Vol. of oxygen

∴ 10 cm³ C₂H₂ will require $\frac{5}{2}$ x 10

= 25 cm³ of oxygen

Given, air contains 20% O₂ by volume

Let $x$ volume of air contain 25 cm³ of O₂

$$\begin{aligned}\Rightarrow \frac{20}{100} \times x = 25 \\[1em] \Rightarrow x = \frac{100}{20} \times 25 \\[1em] \Rightarrow x = 125 \text{ cm}^3\end{aligned}$$

∴ 25 cm³ O₂ is present in 125 cm³ of air.

Hence, total volume of air required is 100 + 125 = 225 cm³

Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litres

$$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} \\ 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} \\ \end{matrix}$$

1 Vol. C₃H₈ produces carbon dioxide = 3 Vol

So, 6 litres C₃H₈ will produce carbon dioxide = 3 x 6 = 18 litres

2 Vol. C₄H₁₀ produces carbon dioxide = 8 Vol

So, 4 litres C₄H₁₀ will produce carbon dioxide = $\frac{8}{2}$ x 4 = 16 litres

Hence, 34 (i.e., 18 + 16) litres of CO₂ is produced.

$$\begin{matrix} \text{C}_2\text{H}_4 & + & 3\text{O}_2 & \longrightarrow & 2\text{CO}_2 + 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 3 \text{ vol.} \\ 11 \text{ lit} & : & 33 \text{ lit} \end{matrix}$$

STP	Given Values
P1 = 760 mm of Hg	P2 = 380 mm of Hg
V1 = x lit	V2 = 33 lit
T1 = 273 K	T2 = 273 + 273 K = 546 K

Using the gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$$\begin{aligned}\frac{760 \times x}{273} = \frac{380 \times 33}{546} \\[0.5em] x = \frac{380 \times 33 \times 273}{546 \times 760 } \\[0.5em] x = \frac{3,423,420}{414,960} \\ \\[0.5em] x = 8.25 \text{ lit}\end{aligned}$$

Hence, volume of oxygen required = 8.25 lit.

Gas	Volume (in litres)	Number of molecules
Chlorine	10	x/2
Nitrogen	20	x
Ammonia	20	x
Sulphur dioxide	5	x/4

Reason — According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain same number of molecules. If 20 lit of nitrogen contains x molecules then 20 lit of ammonia will also contain x molecules. As volume of chlorine is half that of nitrogen so it will contain half the number of molecules of nitrogen i.e., x/2. Similarly, sulphur dioxide will contain x/4 molecules.

(a) Given, 150 cc of gas A contains X molecules. According to Avogadro's law, 150 cc of gas B will also contain X molecules.

So, 75 cc of gas B will contain $\frac{x}{2}$ molecules.

(b) The problem is based on Avogadro's law.

(a) (NH₄)₂PtCl₆

= (2N) + (8H) + (Pt) + (6Cl)

= (2 x 14) + (8 x 1) + 195 + (6 x 35.5)

= 28 + 8 + 195 + 213

= 444 a.m.u.

(b) KClO₃

= (K) + (Cl) + (3O)

= 39 + 35.5 + (3 x 16)

= 39 + 35.5 + 48

= 122.5 a.m.u.

(c) CuSO₄.5H₂O

= (Cu) + (S) + (4O) + 5(2H + O)

= 63.5 + 32 + (4 x 16) + 5[(2 x 1) + 16]

= 63.5 + 32 + 64 + (5 x 18)

= 63.5 + 32 + 64 + 90

= 249.5 a.m.u.

(d) (NH₄)₂Cr₂O₇

= (2N) + (8H) + (2Cr) + (7O)

= (2 x 14) + (8 x 1) + (2 x 51.9) + (7 x 16)

= 28 + 8 + 103.8 + 112

= 251.8 ≈ 252 a.m.u.

(a) Number of moles in 10 g of CaCO₃ —

Mass of 1 mole of CaCO₃ = 40 + 12 + 3(16) = 52 + 48 = 100 g

100 g of CaCO₃ = 1 mole

∴ 10 g of CaCO₃ = $\frac{1 \times 10}{100}$

= 0.1 mole

(b) Weight of 0.2 mole H₂ gas —

1 mole of H₂ weighs = 2 g

∴ 0.2 moles will weigh = $\frac{2 \times 0.2}{1}$ = 0.4 g

(c) No. of molecules in 3.2 g of SO₂ —

Molecular wt. of any substance contain 6 × 10²³ molecules.

Mass of 1 mole of SO₂ is 32 + 2(16) = 32 + 32 = 64 g

64 g of SO₂ contains 6 × 10²³ molecules

∴ 3.2 g of SO₂ contains $\frac{6 \times 10^{23} \times 3.2 }{64}$

= 3 x 10²² molecules.

1 mole of CO₂

Reason —

Weight of H₂O = 2 + 16 = 18 g

Weight of CO₂ = 12 + (2 x 16) = 12 + 32 = 44 g

Weight of NH₃ = 14 + (3 x 1) = 14 + 3 = 17 g

Weight of CO = 12 + 16 = 28 g

As weight of CO₂ is maximum, hence 1 mole of CO₂ will weigh the most.

4 g of NH₃

Reason —

(a) No. of molecules in 4 g of O₂

Molecular wt. of any substance contain 6.022 × 10²³ molecules.

Mass of 1 mole of O₂ is 2(16) = 32 g

32 g of O₂ contains 6.022 × 10²³ molecules

∴ 4 g of O₂ contains $\frac{6.022 \times 10^{23} \times 4}{32}$

= 7.5 x 10²² molecules.

Similarly,

(b) 4 g of NH₃ [14 + 3 = 17g ] contains $\frac{6.022 \times 10^{23} \times 4}{17}$

(c) 4 g of CO [12 + 16 = 28g ] contains $\frac{6.022 \times 10^{23} \times 4}{28}$

(d) 4 g of SO₂ [32 + 32 = 64g ] contains $\frac{6.022 \times 10^{23} \times 4}{64}$

∴ 4g of NH₃ having minimum molecular mass contains maximum molecules.

Note : The fraction with lowest denominator gives the highest value. Hence, by observation we can say that 4 g of NH₃ has maximum number of molecules.

No. of particles in 1 mole = 6.022 × 10²³

∴ No. of particles in 0.1 mole = $\frac{6.022 \times 10^{23} \times 0.1}{1}$

= 6.022 × 10²²

1 mole of H₂SO₄ contains (2 × 6.022 × 10²³) hydrogen atoms

∴ 0.1 mole of H₂SO₄ contains = $\frac{6.022 \times 10^{23} \times 2 \times 0.1}{1}$

= 1.2 × 10²³ atoms of hydrogen

Mass of 1 mole of CaCl₂ = Ca + 2Cl = 40 + (2 x 35.5) = 40 + 71 = 111 g

111 g of CaCl₂ contains 6.022 × 10²³ molecules

∴ 1000 g of CaCl₂ contains $\frac{6.022 \times 10^{23} \times 1000}{111}$

= 5.42 × 10²⁴ molecules

1 mole of aluminium has mass = 27 g

0.2 mole of aluminium has mass

= $\frac{27}{1}$ x 0.2

= 5.4 g

1 mole of HCl has mass = 1 + 35.5 = 36.5 g

0.1 mole of HCl has mass

= $\frac{36.5}{1}$ x 0.1

= 3.65 g

1 mole of H₂O has mass = 2(1) + 16 = 2 + 16 = 18 g

0.2 mole of H₂O has mass

= $\frac{18}{1}$ x 0.2

= 3.6 g

1 mole of CO₂ has mass = 12 + 2(16) = 12 + 32 = 44 g

0.1 mole of CO₂ has mass

= $\frac{44}{1}$ x 0.1

= 4.4 g

5.6 litres of gas at S.T.P. has mass = 12 g

∴ 22.4 litre (molar volume) has mass

= $\frac{12}{5.6}$ x 22.4

= 48 g

Oxygen in 1 mole of CO₂ = 2O = (2 x 16) = 32 g

or we can say, 32 g of oxygen is present in 1 mole of CO₂

∴ 8 gm of O₂ is present in $\frac{1}{32}$ x 8

= 0.25 moles

Molar mass of methane (CH₄) = C + 4H = 12 + 4 = 16 g

16 g of methane = 1 mole

∴ 0.80 g of methane = $\frac{1}{16}$ x 0.80

= 0.05 moles

(a) Number of oxygen atoms in 16 g of atomic oxygen = 6.022 × 10²³ atoms

∴ mass of 1 atom of oxygen

= $\frac{16}{6.022 \times 10^{23}}$

= 2.657 × 10^-23 g

(b) Number of hydrogen atoms in 1 g of atomic hydrogen = 6.022 × 10²³ atoms

∴ Mass of 1 atom of hydrogen

= $\frac{1}{6.022 \times 10^{23}}$

= 1.666 × 10^-24 g

(c) Gram molecular mass of NH₃ = 14 + 3 = 17 g

Number of NH₃ molecules in 17 g of NH₃ = 6.022 × 10²³ molecules

Mass of 6.022 × 10²³ molecules of NH₃ = 17g

∴ Mass of 1 molecule of NH₃ = $\frac{17}{6.022 \times 10^{23}}$

= 2.823 × 10^-23 g

(d) Mass of 6.022 × 10²³ atoms of atomic carbon = 12 g

∴ Mass of 10²² atoms of carbon = $\frac{12}{6.022 \times 10^{23}} \times 10^{22}$

= 0.2 g

(e) Atomic weight of calcium = 40 g

Gram atom = $\frac{\text{Mass of element}}{\text{Atomic mass}}$

Therefore, 0.25 = $\frac{\text{Mass of calcium}}{40}$

Mass of calcium = 40 x 0.25 = 10 g

(a) Vapour density of carbon dioxide is 22 implies that 1 molecule of carbon dioxide is 22 times heavier than 1 molecule of hydrogen.

(b) Vapour density = $\frac{\text{Molecular mass}}{2}$

Molecular mass of chlorine Cl₂ = 2Cl = 2 x 35.5 = 71 g

Substituting in formula;

Vapour density = $\frac{71}{2}$ = 35.5

Hence, vapour density of Chlorine atom is 35.5

1 molecule of Na₂CO₃.10H₂O contains 13 atoms of oxygen

∴ 6.022 × 10²³ molecules (ie., 1 mole) has 13 × 6.022 × 10²³ atoms

∴ 0.1 mole will have atoms = 0.1 × 13 × 6.022 × 10²³

= 7.8 × 10²³ atoms

Atomic mass of Na = 23

23 g of sodium = 1 gram atom of sodium

∴ 4.6 gram of sodium = $\frac{\text{{4.6}}}{\text{{23}}}$

= 0.2 gram atom of sodium

32 g of oxygen = 1 mole

∴ 12 g of oxygen = $\frac{12}{32}$ = $\frac{3}{8}$

= 0.375 mole

1 mole of Sulphur weighs 32 g and contains 6.02 x 10²³ atoms

∴ 3.2 g of Sulphur will contain = $\frac{6.02 \times 10^{23}}{32} \times 3.2$

= 6.02 x 10²² atoms.

6.02 x 10²³ atoms of Ca weighs = 40 g

∴ 6.02 x 10²² atoms of Ca will weigh = $\frac{40}{ 6.02 \times 10^{23}}$ x 6.02 x 10²² = 4 g.

(a) No. of atoms = Moles x 6.022 x 10²³

= 52 × 6.022 x 10²³ = 3.131 × 10²⁵ atoms

(b) 4 amu = 1 atom of He

∴ 52 amu = $\frac{52}{4}$ = 13 atoms of He

(c) Mass of 1 mole of He is 4 g

4 g of He contains 6.022 × 10²³ atoms

∴ 52 g of He contains $\frac{6.022 \times 10^{23}}{4} \times 52$

= 7.828 × 10²⁴ atoms

Molecular mass of Na₂CO₃ = 2Na + C + 3O = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106 g

(i) 106 g of Na₂CO₃ has = 2 × 6.022 × 10²³ atoms of Na

∴ 5.3 g of Na₂CO₃ will have = $\frac{2 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = 6.022 × 10²² atoms of Na

(ii) 106 g of Na₂CO₃ has = 6.022 × 10²³ atoms of carbon

∴ 5.3 g of Na₂CO₃ will have = $\frac{6.022 \times 10^{23} \times 5.3 }{106}$ = 3.01 × 10²² atoms of carbon

(iii) 106 g of Na₂CO₃ has 3 x 6.022 × 10²³ atoms of oxygen

∴ 5.3 g of Na₂CO₃ will have = $\frac{3 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = 9.03 × 10²² atoms of oxygen

Molar mass of urea [CO(NH₂)₂] = 12 + 16 + 2(14 + (2 x 1))

= 28 + 2(16)

= 28 + 32

= 60 g

Molar mass of nitrogen = 2 x 14 = 28 g

60 g urea has mass of nitrogen = 28 g

∴ 5000 g urea will have mass

= $\frac{28 \times 5000 }{60}$

= 2333 g = 2.33 kg

Molar mass of sulphur dioxide (SO₂) = S + 2O = 32 + (2 x 16) = 32 + 32 = 64 g

64 g of sulphur dioxide has volume = 22.4 litre

∴ 320 g of sulphur dioxide will have volume = $\frac{22.4\times 320}{64}$

= 112 litres

Molar mass of C₁₂H₂₂O₁₁ = 12C + 22H + 11O = (12 x 12) + (22 x 1) + (11 x 16) = 144 + 22 + 176 = 342 g

1000 g of sugar costs = ₹40

∴ 342 g of sugar will cost = $\frac{40}{1000}$ x 342 = ₹13.68 per mole

22400 cm³ of CO has mass = 12 + 16 = 28 g

∴ 56 cm³ will have mass = $\frac{28}{22400}$ x 56 = 0.07 g

Molecular wt. of any substance contain 6.022 × 10²³ molecules.

Mass of 1 mole of water is 2H + O = 2 + 16 = 18 g

18 g of H₂O contains 6.022 × 10²³ molecules

∴ 0.09 g of H₂O contains $\frac{6.022 \times 10^{23} \times 0.09 }{18}$

= 3.01 × 10²¹ molecules

(a) Mass of 1 mole of S₈ = 8S = 8 x 32 = 256 g

∴ Moles in 5.12 g of sulphur = $\frac{5.12}{256}$ = 0.02 moles

(b) 1 mole = 6.022 × 10²³ molecules

∴ 0.02 moles will have = 0.02 × 6.022 × 10²³
= 1.2044 × 10²² ≈ 1.2 × 10²² molecules

No. of atoms in 1 molecule of S₈ = 8

∴ No. of atoms in 1.2044 × 10²² molecules = 1.2044 x 10²² × 8

= 9.635 × 10²² molecules

Mass of 1 mole of P₄ = 4P = 4 x 31 = 124 g

124 g of phosphorus (P₄) = 1 mole

∴ 100 g of phosphorus (P₄) = $\frac{1}{124}$ x 100 = 0.806 moles

(a) The mass of 22.4 L of a gas at S.T.P. is equal to it's gram molecular mass.

308 cm³ of chlorine weighs 0.979 g

∴ 22,400 cm³ of chlorine will weigh

= $\frac{0.979}{308}$ × 22400 = 71.2 g

(b) Molar mass of H₂ = 2H = 2 x 1 = 2 g

2g H₂ at 1 atm has volume = 22.4 dm³

∴ 4 g H₂ at 1 atm will have volume 2 x 22.4 = 44.8 dm³

Now, For 4 g H₂

P₁ = 1 atm, V₁ = 44.8 dm³

P₂ = 4 atm, V₂ = ?

Using formula P₁V₁ = P₂V₂

$$\begin{aligned}\text{V}_2 = \frac{\text{P}_1\text{V}_1}{\text{P}_2} \\[1em] \text{V}_2 = \frac{1 \times 44.8}{4} \\[1em] = \bold{11.2} \space \bold{dm^3}\end{aligned}$$

(c) Molar mass of oxygen in carbon dioxide = 2O = 2 x 16 = 32 g

Mass of oxygen in 22.4 litres of CO₂ = 32 g

∴ Mass of oxygen in 2.2 litres of CO₂

= $\frac{32}{22.4}$ x 2.2 = 3.14 g

No. of atoms in 12 g C = 6.022 × 10²³

∴ no. of carbon atoms in 10^-12 g

$\frac{6.022 \times 10^{23}}{12}$ x 10^-12

= 5.019 × 10¹⁰ atoms

Given:

P = 1140 mm Hg

Density = D = 3 g per L

T = 273 °C = 273 + 273 = 546 K

gram molecular mass = ?

At S.T.P., the volume of one mole of any gas is 22.4 L

Volume of unknown gas at S.T.P. = ?

By Charle’s law.

V₁ = 1 L

T₁ = 546 K

T₂ = 273 K

V₂ = ?

$\frac{\text{V}_1}{\text{T}_1}$ = $\frac{\text{V}_2}{\text{T}_2}$

Hence, V₂ = $\frac{1}{546}$ x 273 = 0.5 L

Volume at standard pressure = ?

Apply Boyle’s law.

P₁ = 1140 mm Hg

V₁ = 0.5 L

P₂ = 760 mm Hg

V₂ = ?

P₁ × V₁ = P₂ × V₂

V₂ = $\frac{1140 \times 0.5}{760}$ = 0.75 L

Now,

22.4 L = 1 mole of any gas at S.T.P.,

then 0.75 L = $\frac{0.75}{22.4}$

= 0.0335 moles

The original mass is 3 g

Molecular mass = $\frac{\text{Mass of compound}}{\text{Moles of compound}}$

= $\frac{3}{0.0335 }$ = 89.55 ≈ 89.6 g per mole

Hence, the gram molecular mass of the unknown gas is 89.6g

(a) Equal volumes of any gas, under similar conditions, contain an equal number of molecules.

(b) 44 g of CO₂, occupies 22.4 litres at STP.

(c) The unit of atomic weight is atomic mass unit (a.m.u).

(a) Molecular formula is C₆H₆

∴ Ratio of C and H is 6 : 6

Simple ratio is 1 : 1

Hence, empirical formula = CH

(b) Molecular formula is C₆H₁₈O₃

∴ Ratio of C, H and O is 6 : 18 : 3

Simple ratio is 2 : 6 : 1

Hence, empirical formula = C₂H₆O

(c) Molecular formula is C₂H₂

∴ Ratio of C and H is 2 : 2

Simple ratio is 1 : 1

Hence, empirical formula = CH

(d) Molecular formula is CH₃COOH i.e. C₂H₄O₂

∴ Ratio of C, H and O is 2 : 4 : 2

Simple ratio is 1 : 2 : 1

Hence, empirical formula = CH₂O

Relative molecular mass of CuSO₄.5H₂O

= 64 + 32 + (4×16) + [5(2+16)]

= 96 + 64 + 90 = 250

250 g of CuSO₄.5H₂O contains 90 g of water of crystallisation

∴ 100 g of CuSO₄.5H₂O contains

= $\frac{90}{250}$ x 100 = 36%

Molecular mass of Ca(H₂PO₄)₂

= Ca + 2[2H + P + 4O]

= 40 + 2[2(1) + 31 + 4(16)]

= 40 + 2[2 + 31 + 64]

= 40 + 194

= 234

234 g of Ca(H₂PO₄)₂ contains 62 g of P

∴ 100 g of Ca(H₂PO₄)₂ contains

= $\frac{62}{234}$ x 100 = 26.5%

Molecular mass of KClO₃

= K + Cl + 3O

= 39 + 35.5 + (3 x 16)

= 39 + 35.5 + 48

= 122.5 g

% of K = ?

Since, 122.5 g of KClO₃ contains 39 g of K

∴ 100 g of KClO₃ contains

= $\frac{39}{122.5}$ x 100

= 31.83%

Similarly, 122.5 g of KClO₃ contains 35.5 g of Cl

∴ 100 g of KClO₃ contains

= $\frac{35.5}{122.5}$ x 100

= 28.98%

And, 122.5 g of KClO₃ contains 48 g of O

∴ 100 g of KClO₃ contains

= $\frac{48}{122.5}$ x 100

= 39.18%

Element	% composition	At. mass	Relative no. of atoms	Simplest ratio
Pb	62.5	207	$\frac{62.5 }{207}$ = 0.301	$\frac{0.301}{0.301 }$ = 1
N	8.5	14	$\frac{8.5}{14}$ = 0.607	$\frac{0.607}{0.301 }$ = 2
O	29	16	$\frac{29}{16}$ = 1.81	$\frac{1.81}{0.301 }$ = 6

Hence, Simplest ratio of whole numbers = Pb : N : O = 1 : 2 : 6

Hence, empirical formula is Pb(NO₃)₂.

Atomic wt. of Fe = 56 and O = 16

Molecular mass of Fe₂O₃ = 2Fe + 3O

=(2 x 56) + (3 x 16)

= 112 + 48

= 160 g

Iron present in 80% of Fe₂O₃ = $\frac{112}{160}$ x 80

= 56 g

∴ Mass of iron in 100 g of iron ore = 56 g

Hence, mass of iron present in 10 kg (i.e., 10,000 g) of iron ore = $\frac{56}{100}$ x 10000

= 5600 g = 5.6 kg

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 13

Molecular weight = 2 x V.D. = 2 x 13

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 13}{13} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[CH] = C₂H₂

Similarly,

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 39

Molecular weight = 2 x V.D. = 2 x 39

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 39}{13} = 6\end{aligned}$$

∴ Molecular formula = n[E.F.] = 6[CH] = C₆H₆

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Nitrogen	82.35	14	$\frac{82.35 }{14}$ = 5.88	$\frac{5.88}{5.88}$ = 1
Hydrogen	17.64	1	$\frac{17.64}{1}$ = 17.64	$\frac{17.64}{5.88}$ = 3

Simplest ratio of whole numbers = N : H = 1 : 3

Hence, empirical formula is NH₃

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	54.54	12	$\frac{54.54}{12}$ = 4.545	$\frac{ 4.545 }{2.275 }$ = 1.99 = 2
Hydrogen	9.09	1	$\frac{9.09 }{1}$ = 9.09	$\frac{9.09 }{ 2.275 }$ = 3.99 = 4
Oxygen	36.36	16	$\frac{36.36}{16}$ = 2.275	$\frac{2.275 }{ 2.275 }$ = 1

Simplest ratio of whole numbers = C : H : O = 2 : 4 : 1

Hence, empirical formula is C₂H₄O

Empirical formula weight = 2(12) + 4(1) + 16 = 24 + 4 + 16 = 44

V.D. = 44

Molecular weight = 2 x V.D. = 2 x 44 = 88

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{88}{44} = 2\end{aligned}$$

So, molecular formula = (C₂H₄O)₂ = C₄H₈O₂

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Hydrogen	2.22	1	$\frac{2.22}{1}$ = 2.22	$\frac{2.22 }{2.21}$ = 1
Oxygen	71.19	16	$\frac{71.19}{16}$ = 4.44	$\frac{4.44}{2.21 }$ = 2
Carbon	26.59	12	$\frac{26.59}{12}$ = 2.21	$\frac{ 2.21 }{2.21}$ = 1

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, empirical formula is CHO₂

Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{90}{45} = 2\end{aligned}$$

So, molecular formula = 2(CHO₂) = C₂H₂O₄

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Hydrogen	4.07	1	$\frac{4.07 }{1}$ = 4.07	$\frac{4.07 }{2.01}$ = 2
chlorine	71.65	35.5	$\frac{ 71.65}{35.5}$ = 2.01	$\frac{2.01}{2.01 }$ = 1
Carbon	24.28	12	$\frac{24.28 }{12}$ = 2.02	$\frac{ 2.02 }{2.01}$ = 1

Simplest ratio of whole numbers = H : Cl : C = 2 : 1 : 1

Hence, empirical formula is CH₂Cl

Empirical formula weight = 12 + (2 x 1) + 35.5 = 49.5

molar mass = 98.96

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{98.96}{49.5} = 1.99 = 2\end{aligned}$$

So, molecular formula = 2(CH₂Cl) = C₂H₄Cl₂

(a) Given, hydrocarbon contains 4.8 g of carbon per gram of hydrogen

Gram atom = $\frac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of carbon = $\frac{4.8}{12}$ = 0.4 and

g atom of hydrogen = $\frac{1}{1}$ = 1

(b)

Element	Mass	At. wt.	Gram atoms	Simplest ratio
Hydrogen	1	1	$\frac{1 }{1}$ = 1	$\frac{1 }{0.4 }$ = $\frac{5 }{2}$
Carbon	4.8	12	$\frac{4.8 }{12}$ = 0.4	$\frac{ 0.4 }{0.4 }$ = 1

Simplest ratio of whole numbers = H : C = $\frac{5 }{2}$ : 1 = 5 : 2

Hence, empirical formula is C₂H₅

(c) Empirical formula weight = (2 x 12) + (5 x 1) = 24 + 5 = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29 = 58

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{58}{29} = 2\end{aligned}$$

So, molecular formula = 2(C₂H₅) = C₄H₁₀

Gram atom = $\frac{\text{Mass of element}}{\text{Atomic mass}}$

g atom of silicon = 0.2 = $\frac{\text{Mass of silicon}}{28}$

∴ Mass of silicon = 5.6 g and

Mass of chlorine = 21.3 g

Element	Mass	At. wt.	gram atoms	Simplest ratio
Silicon	5.6	28	$\frac{5.6 }{28}$ = 0.2	$\frac{0.2 }{0.2}$ = 1
Chlorine	21.3	35.5	$\frac{ 21.3}{35.5}$ = 0.6	$\frac{ 0.6 }{0.2}$ = 3

Simplest ratio of whole numbers = Si : Cl = 1 : 3

Hence, empirical formula is SiCl₃

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	82.76	12	$\frac{82.76}{12}$ = 6.89	$\frac{6.89}{6.89}$ = 1
Hydrogen	17.24	1	$\frac{17.24}{1}$ = 17.24	$\frac{17.24}{6.89}$ = $\frac{5 }{2}$

Simplest ratio of whole numbers = C : H = 1 : $\frac{5 }{2}$ = 2 : 5

Hence, empirical formula is C₂H₅

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 29}{29} =2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[C₂H₅] = C₄H₁₀

(a) Gram atom = $\frac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of magnesium = $\frac{18}{24}$ = $\frac{3}{4}$

Hence, $\frac{3}{4}$ gram atoms of magnesium are equal to 18g of magnesium.

(b) g atom of nitrogen = $\frac{7}{14}$ = $\frac{1}{2}$

Hence, $\frac{1}{2}$ gram atoms of nitrogen are equal to 7g of nitrogen.

(c) simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen

= $\frac{\frac{3}{4}}{\frac{1}{2}}$ = $\frac{3}{2}$ = magnesium : nitrogen

So, the formula is Mg₃N₂

Barium chloride = BaCl₂.xH₂O

Molecular weight of BaCl₂.xH₂O = Ba + 2Cl + x(2H + O)

= 137 + (2 x 35.5) + x(2+16)

= 137 + (2 x 35.5) + x(2+16)

= 137 + 71 + 18x

= (208 + 18 x)

(208 + 18 x) contains 14.8% of water of crystallisation in BaCl2.x H₂O

∴ 14.8% of (208 + 18 x) = 18x

$\Big[\frac{14.8}{100}\Big]$ x [208 + 18 x] = 18x

[0.148 x 208 ] + [0.148 x 18x] = 18x

30.784 = 18x - [0.148 x 18x]

30.784 = 18x - 2.664x

30.784 = 15.336x

x = $\frac{30.784}{15.336}$ = 2

Hence, Barium chloride crystals contain 2 molecules of water of crystallisation per molecule.

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Nitrogen	87.5	14	$\frac{87.5 }{14}$ = 6.25	$\frac{6.25}{6.25 }$ = 1
Hydrogen	12.5	1	$\frac{12.5}{1}$ = 12.5	$\frac{12.5}{6.25 }$ = 2

Simplest ratio of whole numbers = N : H = 1 : 2

Hence, empirical formula is NH₂

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Oxygen	51.42	16	$\frac{51.42 }{16}$ = 3.21	$\frac{3.21}{3.21}$ = 1
Hydrogen	6.48	1	$\frac{6.48}{1}$ = 6.48	$\frac{6.48}{3.21}$ = 2
Carbon	42.1	12	$\frac{42.1}{12}$ = 3.50	$\frac{3.50}{3.21}$ = 1

Simplest ratio of whole numbers = O : H : C = 1 : 2 : 1

Hence, empirical formula is CH₂O

Since the compound has 12 atoms of carbon, so the formula is C₁₂H₂₄O₁₂.

Empirical formula = AB₂

Empirical formula weight = V.D.

Molecular weight = 2 x V.D.

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \frac{\text{2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[AB₂] = A₂B₄

Given,

Empirical formula = AB

V.D. = 3 x Empirical formula weight

Hence, Empirical formula weight = $\frac{\text{V.D.}}{3}$

and we know, Molecular weight = 2 x V.D.

Substituting in the formula for n we get,

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \frac{\text{ 2 x V.D.}}{\frac{\text{V.D.}}{3}} \\[0.5em] \text{n} = \frac{\text{ 3 x 2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 6\end{aligned}$$

∴ Molecular formula = n[E.F.] = 6[AB] = A₆B₆

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Metal A	6.25	207	$\frac{6.25 }{207}$ = 0.03	$\frac{0.03}{0.03}$ = 1
Non-metal B	4.22	35.5	$\frac{4.22}{35.5}$ = 0.11	$\frac{0.11}{0.03}$ = 3.96 = 4

Simplest ratio of whole numbers = A : B = 1 : 4

Hence, empirical formula is AB₄

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Zn	22.65	65	$\frac{22.65}{65}$ = 0.3484	$\frac{0.3484}{0.3484}$ = 1
S	11.15	32	$\frac{11.15}{32}$ = 0.3484	$\frac{0.3484}{0.3484}$ = 1
O	61.32	16	$\frac{61.32 }{16}$ = 3.832	$\frac{3.832}{0.3484}$ = 10.99 = 11
H	4.88	1	$\frac{4.88}{1}$ = 4.88	$\frac{4.88 }{0.3484}$ = 14

Simplest ratio of whole numbers = Zn : S : O : H = 1 : 1 : 11 : 14

Hence, empirical formula is ZnSO₁₁H₁₄

Molecular weight = 287

Empirical formula weight = 65 + 32 + 11(16) + 14(1) = 65 + 32 + 176 + 14 = 287

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{287}{287} = 1\end{aligned}$$

Molecular formula = n[E.F.] = 1[ZnSO₁₁H₁₄] = ZnSO₁₁H₁₄

Since all the hydrogen in the compound is in combination with oxygen as water of crystallization .

Therefore, 14 atoms of hydrogen and 7 atoms of oxygen = 7H₂O and hence, 4 atoms of oxygen remain.

∴ Molecular formula is ZnSO₄.7H₂O.

(a)

$$\begin{matrix} \text{CaCO}_3 & + \space 2\text{HNO}_3 \longrightarrow & \text{Ca(NO}_3)_2 & + \space \text{H}_2\text{O} \space + \text{CO}_2 \\ 40 + 12 + 3(16) & & 40 + 2(14) + 6(16) \\ = 40 + 12 + 48 & & 40 + 28 + 96 \\ = 100 \text{ g} & & 164 \text{ g} \end{matrix}$$

100 g of CaCO₃ produces = 164 g of Ca(NO₃)₂

∴ 15 g CaCO₃ will produce = $\frac{164}{100}$ x 15

= 24.6 g

Hence, mass of anhydrous calcium nitrate formed = 24.6 g

(b) 100 g of CaCO₃ produces = 22.4 litres of carbon dioxide

∴ 15 g of CaCO₃ will produce = $\frac{22.4}{100}$ x 15

= 3.36 litres of CO₂

(a) $$\begin{matrix} 2\text{NH}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{(NH}_4)_2\text{SO}_4 \\ 2[14 + 3(1)] & & 2(1) + 32 + 4(16) & & 2[14 + 4(1)] + 32 + 4(16) \\ = (2 \times 17) & & = 2 + 32 + 64 & & = 36 + 32 + 64 \\ = 34 \text{ g} & & = 98 \text{ g} & & = 132 \text{ g} \\ 2\text{ mole} \end{matrix}$$

132 g ammonium sulphate is produced by 34 g of NH₃

∴66 g ammonium sulphate is produced by $\frac{34}{132}$ x 66 = 17 g of NH₃

Hence, 17g of NH₃ is required.

(b) 132 g ammonium sulphate uses 2 x 22.4 L of gas

∴ 66 g of ammonium sulphate will use $\frac{2 \times 22.4}{132}$ x 66 = 22.4 litres

(c) For 132 g ammonium sulphate 98 g of acid is required

∴ For 66 g ammonium sulphate $\frac{98}{132}$ x 66 = 49 g

Hence, 49g of acid is required.

(a)

$$\begin{matrix} \text{Pb}_3\text{O}_4 & + &8\text{HCl} & \longrightarrow \\ 3(207) + 4(16) & & 8[1 + 35.5] & \\ = 621 + 64 & & = 8(36.5) & \\ = 685 \text{ g} & & = 292 \text{ g} & \\ & & & \\ 3\text{PbCl}_2 & + & 4\text{H}_2\text{O} & + & \text{Cl}_2 \\ 3[207 + 2(35.5)] &&&& 2(35.5) \\ = 3[207 + 71] &&&& = 71\text{g} \\ = 834 \text{ g} \end{matrix}$$

685 g of Pb₃O₄ gives = 834 g of PbCl₂

∴ 6.85 g of Pb₃O₄ will give

= $\frac{834}{685}$ x 6.85 = 8.34 g

(b) 685g of Pb₃O₄ gives = 71g of Cl₂

∴ 6.85 g of Pb₃O₄ will give

= $\frac{71}{685}$ x 6.85 = 0.71 g of Cl₂

(c) 685 g of Pb₃O₄ produces 22.4 L of Cl₂

∴ 6.85 g of Pb₃O₄ will produce

$\frac{22.4}{685}$ x 6.85 = 0.224 L of Cl₂

$$\begin{matrix} \text{KNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{KHSO}_4 & + \text{HNO}_3 \\ 39 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 39 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 101 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$$

63 g of HNO₃ is formed by 101 g of KNO₃

∴ 126000 g of HNO₃ is formed by $\frac{101}{63}$ x 126000
= 202000 g = 202 kg

Similarly,

$$\begin{matrix} \text{NaNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{NaHSO}_4 & + \text{HNO}_3 \\ 23 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 23 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 85 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$$

63 g of HNO₃ is formed by 85 g of NaNO₃

∴ 126000 g of HNO₃ is formed by $\frac{85}{63}$ x 126000
= 170000 g = 170 kg

So, a smaller mass of NaNO₃ is required.

(a) Given,

$$\begin{matrix} \text{CaCO}_3 & + &2\text{HCl} & \longrightarrow & \text{CaCl}_2 & + \text{H}_2\text{O} + &\text{CO}_2 \\ 40 + 12 + 3(16) & & 2[1 + 35.5] & && & 1\text{ mole} \\ = 40 + 12 + 48 && = 73 \text{ g} \\ = 100 \text{ g} \\ \end{matrix}$$

First convert the volume of carbon dioxide to STP:

V₁ = 2 L

T₁ = 27 + 273 K = 300 K

T₂ = 273 K

V₂ = ?

Using formula:

$\frac{\text{V}_1}{\text{T}_1}$ = $\frac{\text{V}_2}{\text{T}_2}$

Substituting in the formula,

$\frac{2}{300}$ = $\frac{\text{V}_2}{273}$

V₂ = $\frac{2}{300}$ x 273 = 1.82 L

As, 22.4 L of carbon dioxide is obtained using 100 g CaCO₃

∴ 1.82 L of carbon dioxide is obtained from $\frac{100}{22.4}$ x 1.82

= 8.125 g of CaCO₃

(b) Similarly, 22.4 L of carbon dioxide is obtained using 73 g of acid

∴ 1.82 L of carbon dioxide is obtained from $\frac{73}{22.4}$ x 1.82

= 5.93 g of acid

$$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow 2\text{H}_2 \space + & \text{O}_2 \\ 2(2 + 16) & & 2(16) \\ 36 \text{g} & & 32 \text{g} \\ \end{matrix}$$

36 g of water produces 32 g of O₂

∴ 18 g of water will produced

= $\frac{32}{36}$ x 18 = 16 g of O₂

$$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow 2\text{H}_2 \space + & \text{O}_2 \\ 2\text{ mole} & & 1\text{ mole} \\ \end{matrix}$$

2 moles of water produces 1 mole of oxygen

∴ 1 mole of water will produce $\frac{1}{2} \times 1$ = 0.5 moles of O₂

1 mole of O₂ occupies 22.4 L volume

∴ 0.5 moles will occupy = 22.4 × 0.5

= 11.2 L

$$\begin{matrix} 2\text{Na}_2\text{O}_2 & + \space 2\text{H}_2\text{O} \longrightarrow & 4\text{NaOH}& & + & \text{O}_2 \\ 2[2(23) + 2(16)] & & 4(23 + 16 + 1) & & & 1 \text{mole} \\ 156 \text{g} & & 160\text{g}& & &32 \text{g} \\ \end{matrix}$$

(a) 156 g of sodium peroxide produces 160 g of sodium hydroxide

∴ 1.56 g of sodium peroxide will produce $\frac{160}{156}$ x 1.56

= 1.6 g of sodium hydroxide

(b) 156 g of sodium peroxide produces 22.4 L of oxygen

∴ 1.56 g of sodium peroxide will produce $\frac{22.4}{156}$ x 1.56

= 0.224 L

Converting L to cm³

As 1 L = 1000 cm³

So, 0.224 L = 224 cm³

(c) 156 g of sodium peroxide produces 32 g of oxygen

∴ 1.56 g of sodium peroxide will produce $\frac{32}{156}$ x 1.56 = 0.32 g

$$\begin{matrix} 2\text{NH}_4\text{Cl} & + \text{ Ca(OH)}_2 \longrightarrow \text{CaCl}_2 \space + \space 2\text{H}_2\text{O} \space + & 2\text{NH}_3 \\ 2[14 + 4(1) + 35.5] & & 2[14 + 3(1)] \\ 107 \text{g} & &34\text{g} \\ & &2\text{ mole} \\ \end{matrix}$$

(a) 107 g NH₄Cl gives 34 g of NH₃

∴ 21.4 g NH₄Cl will give $\frac{34}{107}$ x 21.4

= 6.8 g of NH₃

(b) Volume of ammonia produced by 107 g NH₄Cl = 2 x 22.4 L

∴ Volume of ammonia produced by 21.4 g NH₄Cl = $\frac{2 \times 22.4}{107}$ x 21.4

= 8.96 L

$$\begin{matrix} \text{Al}_4\text{C}_3 & + & 12\text{H}_2\text{O} & \longrightarrow & 4\text{Al (OH)}_3 & + & 3\text{CH}_4 \\ 4(27) + 3(12) & & & & 4(78) & & 3(22.4) \\ = 144 \text{ g} & & & & = 312 \text{ g} & & = 67.2 \text{ lit.} \\ \end{matrix}$$

144 g of aluminium carbide forms 312 g of aluminium hydroxide.

∴ 12 g of aluminium carbide will form $\frac{312}{144}$ x 12 = 26 g of aluminium hydroxide

Hence, 26 g of aluminium hydroxide is formed.

(ii) 144 g of aluminium carbide forms 67.2 lit of methane.

∴ 12 g of aluminium carbide will form $\frac{67.2}{144}$ x 12 = 5.6 lit.

Hence, vol. of methane obtained = 5.6 L

$$\begin{matrix} \text{MnO}_2 & + & 4\text{HCl}& \longrightarrow & \text{MnCl}_2 & + & 2\text{H}_2\text{O} & + & \text{Cl}_2 \\ 1 \text{ mole} && 4 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 55 +2(16) && 4[1 + 35.5] & & 55 + 2(35.5) & & &&2(35.5) \\ = 87 \text{ g} & & = 146 \text{ g} & & = 126 \text{ g} & & & & = 71\text{ g} \\ \end{matrix}$$

(a) 1 mole of MnO₂ weighs 87 g

∴ 0.02 mole will weigh $\frac{87}{1}$ x 0.02 = 1.74 g

(b) 1 mole MnO₂ gives 1 mole of MnCl₂

∴ 0.02 mole MnO₂ will give 0.02 mole of MnCl₂

(c) As, 1 mole MnCl₂ weighs 126 g

∴ 0.02 mole MnCl₂ will weigh $\frac{126}{1}$ x 0.02 = 2.52 g

(d) 1 mole MnO₂ gives 1 mole of Cl₂

∴ 0.02 mole MnO₂will form 0.02 moles of Cl₂

(e) 1 mole of Cl₂ weighs 71 g

∴ 0.02 mole will weigh $\frac{71}{1}$ x 0.02 = 1.42 g

(f) 1 mole of chlorine gas has volume 22.4 dm³

∴ 0.02 mole will have volume $\frac{22.4}{1}$ x 0.02 = 0.448 dm³

(g) 1 mole MnO₂ requires 4 moles of HCl

∴ 0.02 mole MnO₂ will require $\frac{4}{1}$ x 0.02 = 0.08 mole

(e) Mass of 1 mole of HCl = 36.5 g

∴ Mass of 0.08 mole = 0.08 × 36.5 = 2.92 g

$$\begin{matrix} \text{N}_2 & + & 3\text{H}_2& \longrightarrow & 2\text{NH}_3 \\ 2(14) && 6(1) & & 2[14 + 3(1)] \\ 28 \text{g} & & 6 \text{g} & & 34 \text{ g} \\ \end{matrix}$$

(a) 28 g of nitrogen requires 6 g of hydrogen

∴ 2000 g of nitrogen requires $\frac{6}{28}$ x 2000

= 428.57 g of hydrogen.

So mass of hydrogen left unreacted = 1000 - 428.57 = 571.42 g

571.42 g of hydrogen is left unreacted.

(b) 28 g of nitrogen forms 34 g NH₃

∴ 2000 g of nitrogen forms $\frac{34}{28}$ x 2000

= 2428.57 g of NH₃