The balanced reaction taking place is given as,

\(\displaystyle\Rightarrow{4}{N}{H}{3}{\left({g}\right)}+{5}{O}{2}{\left({g}\right)}\Rightarrow{4}{N}{O}{\left({g}\right)}+{6}{H}{2}{O}{\left({g}\right)}\)

According to ideal gas law,

\(\displaystyle{P}{V}={n}{R}{T}\)

where \(\displaystyle{P}=\) pressure in atm

\(\displaystyle{V}=\) volume in L

\(\displaystyle{n}=\) moles

\(\displaystyle{R}=\text{gas constant}={0.0821}\ {a}{t}{m}.\frac{{L}}{{K}}.{m}{o}{l}\)

\(\displaystyle{T}=\text{temperature in K}={T}\in{C}^{{\circ}}+{273}\)

Step 2

a) For \(\displaystyle{N}{H}{3},{P}={1.50}\ {a}{t}{m},{V}={12.5}{L}\ {\quad\text{and}\quad}\ {T}={298}{K}\)

Hence substituting the values we get,

\(\displaystyle\Rightarrow{1.50}\times{12.5}={n}\times{0.0821}\times{298}\)

\(\displaystyle\Rightarrow{n}=\text{moles of NH3 taken}={0.7664}\ {m}{o}{l}\) approx.

b) For \(\displaystyle{O}{2},{P}={1.10}\ {a}{t}{m},{V}={18.9}{L}{\quad\text{and}\quad}{T}={323}{K}\)

Hence substituting the values we get,

\(\displaystyle\Rightarrow{1.10}\times{18.9}={n}\times{0.0821}\times{323}\)

\(\displaystyle\Rightarrow{n}=\text{moles of O2 taken}={0.784}\ {m}{o}{l}\) approx.

Step 3

\(\displaystyle\Rightarrow{4}\ {N}{H}{3}{\left({g}\right)}+{5}{O}{2}{\left({g}\right)}\Rightarrow{4}\ {N}{O}{\left({g}\right)}+{6}{H}{2}{O}{\left({g}\right)}\)

From the above balanced reaction we can see that 5 moles of O2 is required to react with 4 moles of NH3

Hence moles of O2 required \(\displaystyle=\text{moles of NH3 taken}\times{\frac{{{5}}}{{{4}}}}={0.7664}\times{\frac{{{5}}}{{{4}}}}={0.958}\ {m}{o}{l}\)

Since moles of O2 taken is less than required. Hence O2 is the limiting reagent and will react completely.

From the above reaction we can see that 5 moles of O2 is reacting to produce 6 moles of H2O.

Hence moles of H2O that can be produced \(\displaystyle=\text{moles of O2 taken}\times{\frac{{{6}}}{{{5}}}}={0.784}\times{\frac{{{6}}}{{{5}}}}={0.9408}\ {m}{o}{l}\)

Molar mass of \(\displaystyle{H}{2}{O}=\text{Atomic mass of H}\times{2}+\text{Atomic mass of O}={1}\times{2}+{16}={18}\frac{{g}}{{m}}{o}{l}\)

Hence mass of H2O that can be produced \(\displaystyle=\text{moles}\times\text{molar mass}={0.9804}\times{18}={16.9}{g}\) approx.

Hence the answer is 16.9 g.