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Ammonia and oxygen react at high temperature as in reaction, $4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_2 \mathrm{O}_{(g)}$ If rate of formation of $\mathrm{NO}$ is $3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}$. Calculate the rate of formation of water.
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$5.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}$
$-\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{d} t}=+\frac{1}{4} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{d} t}=\frac{1}{6} \frac{\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}\right]}{\mathrm{d} t}$
$\begin{aligned} & \frac{\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}\right]}{\mathrm{d} t}=\frac{6}{4} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{d} t}=\frac{3}{2} \times 3.6 \times 10^{-3} \\ & =5.4 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}-\mathrm{sec}}\end{aligned}$
$\begin{aligned} & \frac{\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}\right]}{\mathrm{d} t}=\frac{6}{4} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{d} t}=\frac{3}{2} \times 3.6 \times 10^{-3} \\ & =5.4 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}-\mathrm{sec}}\end{aligned}$
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