JEE Main 2018 — Some Basic Concepts of Chemistry Question with Solution

(a) Molar mass of Ba ${\left({\mathrm{N}}_3\right)}_2(\mathrm{s})=221\mathrm{g}/\mathrm{mol}$ 1 mole of $\mathrm{Ba}{\left({\mathrm{N}}_3\right)}_2(\mathrm{s})$ will give 3 moles of ${\mathrm{N}}_2$ hence $\frac{1\mathrm{g}}{221\mathrm{g}/\mathrm{mol}}$ moles of $\mathrm{Ba}{\left({\mathrm{N}}_3\right)}_2(\mathrm{s})$ will give $3\times \frac{1}{221}$ $=0.014$ moles of ${\mathrm{N}}_2$ (b) Molar mass of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=252\mathrm{g}/\mathrm{mol}$. 1 mole of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ will give 1 mole of ${\mathrm{N}}_2$ hence $\frac{1\mathrm{g}}{252\mathrm{g}/\mathrm{mol}}$ moles of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ will give $1\times \frac{1}{252}=0.0039$ moles of ${\mathrm{N}}_2$ (c) Molar mass of ${\mathrm{NH}}_3=17\mathrm{g}/\mathrm{mol}$. 2 mole of ${\mathrm{NH}}_3$ will give 1 mole of ${\mathrm{N}}_2$ hence $\frac{1\mathrm{g}}{17\mathrm{g}/\mathrm{mol}}$ moles of ${\mathrm{NH}}_3$ will give $$\frac{1}{2\times 17}=0.0297\text{ moles of }{\mathrm{N}}_2\text{. }$$ (d) Molar mass of ${\mathrm{NH}}_4{\mathrm{NO}}_3=80\mathrm{g}/\mathrm{mol}$. 1 mole of ${\mathrm{NH}}_4{\mathrm{NO}}_3$ will give 1 mole of ${\mathrm{N}}_2$ hence $\frac{1\mathrm{g}}{80\mathrm{g}/\mathrm{mol}}$ moles ${\mathrm{NH}}_4{\mathrm{NO}}_3$ will give $$1\times \frac{1}{80}=0.0125\text{ moles of }{\mathrm{N}}_2$$ Hence Thermal decomposition of ${\mathrm{NH}}_3$ will produce maximum amount of ${\mathrm{N}}_2$.