JEE Main 2018 — Chemistry Physical Chemistry

(a) Molar mass of Ba $\left(\mathrm{N}_3\right)_2(\mathrm{~s})=221 \mathrm{~g} / \mathrm{mol}$(N3​)2​( s)=221 g/mol 1 mole of $\mathrm{Ba}\left(\mathrm{N}_3\right)_2(\mathrm{~s})$Ba(N3​)2​( s) will give 3 moles of $\mathrm{N}_2$N2​ hence $\frac{1 \mathrm{~g}}{221 \mathrm{~g} / \mathrm{mol}}$221 g/mol1 g​ moles of $\mathrm{Ba}\left(\mathrm{N}_3\right)_2(\mathrm{~s})$Ba(N3​)2​( s) will give $3 \times \frac{1}{221}$3×2211​ $=0.014$=0.014 moles of $\mathrm{N}_2$N2​ (b) Molar mass of $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7=252 \mathrm{~g} / \mathrm{mol}$(NH4​)2​Cr2​O7​=252 g/mol. 1 mole of $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$(NH4​)2​Cr2​O7​ will give 1 mole of $\mathrm{N}_2$N2​ hence $\frac{1 \mathrm{~g}}{252 \mathrm{~g} / \mathrm{mol}}$252 g/mol1 g​ moles of $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$(NH4​)2​Cr2​O7​ will give $1 \times \frac{1}{252}=0.0039$1×2521​=0.0039 moles of $\mathrm{N}_2$N2​ (c) Molar mass of $\mathrm{NH}_3=17 \mathrm{~g} / \mathrm{mol}$NH3​=17 g/mol. 2 mole of $\mathrm{NH}_3$NH3​ will give 1 mole of $\mathrm{N}_2$N2​ hence $\frac{1 \mathrm{~g}}{17 \mathrm{~g} / \mathrm{mol}}$17 g/mol1 g​ moles of $\mathrm{NH}_3$NH3​ will give $$\frac{1}{2 \times 17}=0.0297 \text { moles of } \mathrm{N}_2 \text {. }$$2×171​=0.0297 moles of N2​.  (d) Molar mass of $\mathrm{NH}_4 \mathrm{NO}_3=80 \mathrm{~g} / \mathrm{mol}$NH4​NO3​=80 g/mol. 1 mole of $\mathrm{NH}_4 \mathrm{NO}_3$NH4​NO3​ will give 1 mole of $\mathrm{N}_2$N2​ hence $\frac{1 \mathrm{~g}}{80 \mathrm{~g} / \mathrm{mol}}$80 g/mol1 g​ moles $\mathrm{NH}_4 \mathrm{NO}_3$NH4​NO3​ will give $$1 \times \frac{1}{80}=0.0125 \text { moles of } \mathrm{N}_2$$1×801​=0.0125 moles of N2​ Hence Thermal decomposition of $\mathrm{NH}_3$NH3​ will produce maximum amount of $\mathrm{N}_2$N2​.