Chemistry - Stoichiometry Question with Solution | TestHub

Step-1: ${\mathrm{NH}}_3(\mathrm{g})+\frac{5}{4}{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{NO}}_{}(\mathrm{g})+\frac{3}{2}{H}_{2}O\left(l\right)$

x x z

${\mathrm{NO}}_{}(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(g)\to {\mathrm{NO}}_2(\mathrm{g})$

x x

$3{\mathrm{NO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(l)\to 2H{\mathrm{NO}}_3(\mathrm{aq})+NO\left(g\right)$

x 2/3x x/3

Step-2: $\mathrm{NO}(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{NO}}_2(\mathrm{g})$

x/3 x/3

$3{\mathrm{NO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\ell )\to 2{\mathrm{HNO}}_3(\mathrm{aq})+\mathrm{NO}(\mathrm{g})$

x/3 $\frac{2}{3}\times \frac{x}{3}$

$\frac{1}{3}\times \frac{x}{3}$

Step-3: $\mathrm{NO}(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{NO}}_2(\mathrm{g})$

$$3{\mathrm{NO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\ell )\to 2{\mathrm{HNO}}_3(\mathrm{aq})+\mathrm{NO}(\mathrm{g})$$

Total moles of ${\mathrm{HNO}}_3$ are calculated as:

$$\frac{2x}{3}+\frac{\frac{2}{3}\times \frac{x}{9}}{9}+\frac{2x}{27}+\dots =\frac{\frac{2x}{3}}{1-\frac{1}{3}}=x$$

The number of moles of ${\mathrm{HNO}}_3$ is:

$$n_{{\mathrm{HNO}}_3}=\frac{378}{63}\times 10^3=6\times 10^3$$

Therefore, the number of moles of ${\mathrm{NH}}_3$ is:

$$n_{{\mathrm{NH}}_3}=6\times 10^3$$

The weight of ${\mathrm{NH}}_3$ is:

$$weightof{\mathrm{NH}}_3=6\times 10^3\times 17=102\times 10^{3}gm=102kg$$