The reaction between concentrated sulphuric acid and magnesium can be represented by the equation given below:
Mg + 2H2SO4 ⟶ MgSO4 + 2H2O + SO2
If 60 g of magnesium is used in the reaction, calculate the following:
(a) The mass of sulphuric acid needed for the reaction.
(b) The volume of sulphur dioxide gas liberated at S.T.P.
[Atomic weight: Mg=24, H=1, S=32, O=16]
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$$\begin{matrix} \text{Mg} & + & 2\text{H}_2 \text{SO}_4 & \longrightarrow & \text{MgSO}_4 & + & 2\text{H}_2\text{O} & + & \text{SO}_2 \\ 1 \text{ mole} && 2 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 24 && 2[2(1) + 32 + 4(16)] & & 24 + 32 + 4(16) & & 2[2(1) + 16] && 32 + 2(16) \\ = 24 \text{ g} & & = 196 \text{ g} & & = 120 \text{ g} & & 36 \text{ g} & & = 64\text{ g} \\ \end{matrix}$$
(a) 24 g of Mg reacts with 196 g of H2SO4
∴ 60 g of Mg reacts with $\frac{196}{24}$ x 60 = 490 g of H2SO4
Hence, 490 g of H2SO4 is required.
(b) 24 g of Mg gives 22.4 L volume of SO2
∴ 60 g of Mg gives $\frac{22.4}{24}$ x 60 = 56 L
Hence, the volume of sulphur dioxide gas liberated at S.T.P. is 56 L.
Thus, 490 g of sulphuric acid are needed and 56 L of SO2 will be liberated at S.T.P.