# Determine the mass of water formed when 12.5 L NH3

Determine the mass of water formed when 12.5 L NH3 (at 298 K and 1.50 atm) is reacted with 18.9 L of O2 (at 323 K and 1.1 atm).
$$\displaystyle{4}{N}{H}{3}{\left({g}\right)}+{5}{O}{2}{\left({g}\right)}\Rightarrow{4}{N}{O}{\left({g}\right)}{6}{H}{2}{O}{\left({g}\right)}$$

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Cheryl King
Step 1
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.
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Mary Herrera

The mass of water formed is
calculation
Use the ideal gas equation to calculate the moles of NH3 and O2
that is $$\displaystyle{P}{v}={n}{R}{T}$$
where; $$\displaystyle{P}=$$ pressure,
$$\displaystyle{V}=$$ volume,
$$\displaystyle{n}=$$ number of moles,
$$\displaystyle{R}=\text{gas constant}={0.0821}{l}.{a}{t}\frac{{m}}{{m}}{o}{l}.{K}$$
make n the formula of the subject by diving both side by RT
$$\displaystyle{n}={\frac{{{P}{V}}}{{{R}{T}}}}$$
The moles of NH3
$$\displaystyle{n}={\frac{{{1.50}{a}{t}{m}\times{12.5}{L}}}{{{0.0821}{L}.{a}{t}\frac{{m}}{{m}}{o}{l}.{k}\times{268}{K}}}}={0.766}\ {m}{o}le{s}$$
The moles of O2
$$\displaystyle={\frac{{{1.1}{a}{t}{m}\times{18.9}{L}}}{{{0.0821}{L}.{a}{t}\frac{{m}}{{m}}{o}{l}.{k}\times{323}{K}}}}={0.784}\ {m}{o}le{s}$$
write the reaction between NH3 and O2
$$\displaystyle{4}{N}{H}{3}+{5}{O}{2}\Rightarrow{4}\ {N}{o}+{6}{H}{2}{O}$$
from equation above 0.766 moles of NH3 reacted to produce
$$\displaystyle{0.766}\times{\frac{{{6}}}{{{4}}}}={1.149}\ {m}{o}le{s}\ {o}{f}\ {H}{2}{O}$$
$$\displaystyle{0.784}\text{moles of O2 reacted to produce}{0.784}\times{\frac{{{6}}}{{{5}}}}={0.9408}\text{moles of H20}$$
since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced $$\displaystyle={0.9408}$$ moles
mass of $$\displaystyle{H}{2}{O}=\text{moles}\times\text{molar mass}$$
from periodic table the molar mass of $$\displaystyle{H}{2}{O}={\left({1}\times{2}\right)}+{16}={18}\frac{{g}}{{m}}{o}{l}$$
$$\displaystyle{m}{a}{s}{s}={18}\frac{{g}}{{m}}{o}{l}\times{0.9408}{m}{o}le{s}={16.93}{g}{r}{a}{m}{s}$$