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10 moles of a mixture of gydogen and oxygen gases at a pressure of $1 \mathrm{~atm}$ at constant volume and temperature, react to form $3.6 \mathrm{~g}$ of liquid water. The pressure of the resulting mixture will be closest to
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The correct answer is:
$0.97$ atm
$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I)$
$0.2$ mole $0.1$ mole $0.2$ mole
moles of gas remaining $=9.7$ at constant $(T) \&(V)$
$\begin{array}{l}
\frac{n_{1}}{n_{2}}=\frac{p_{1}}{p_{2}} \\
\frac{10}{9.7}=\frac{1}{p_{2}} \& p_{2}=0.97
\end{array}$
$0.2$ mole $0.1$ mole $0.2$ mole
moles of gas remaining $=9.7$ at constant $(T) \&(V)$
$\begin{array}{l}
\frac{n_{1}}{n_{2}}=\frac{p_{1}}{p_{2}} \\
\frac{10}{9.7}=\frac{1}{p_{2}} \& p_{2}=0.97
\end{array}$
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