Solid ammonium dichromate decomposes as under:
(NH4)2Cr2O7 ⟶ N2 + Cr2O3 + 4H2O
If 126 g of ammonium dichromate decomposes, calculate:
(a) the number of moles of ammonium dichromate that undergoes decomposition.
(b) the mass of chromic oxide formed at the same time.
(c) the volume of nitrogen gas evolved at STP.
[At. Wt: N=14, Cr=52, O=16, H=1]
Show SolutionHide Solution↓
$$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$$
(a) 252 g of (NH4)2Cr2O7 = 1 mole
∴ 126 g of (NH4)2Cr2O7 = $\frac{1}{252}$ x 126 = 0.5 moles
Hence, no. of moles = 0.5 moles
(b) 252 g of (NH4)2Cr2O7 gives 152 g of Cr2O3
126 g of (NH4)2Cr2O7 will give $\frac{152 \times 126}{252}$ = 76 g of Cr2O3
Hence, mass in gms of Cr2O3 formed = 76 g.
(c) 252 g of (NH4)2Cr2O7 produces 22.4 lit of N2
126 g of (NH4)2Cr2O7 will produce $\frac{22.4 \times 126}{252}$ = 11.2 lit of N2
Hence, volume of N₂ evolved at s.t.p = 11.2 lit.