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$10 \mathrm{~g}$ of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be
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Verified Answer
The correct answer is:
$4 \mathrm{~mol}$
Key Idea
(i) $\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}$
(ii) Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount)
$$
\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 1 \mathrm{~mol} \quad \frac{1}{2} \mathrm{~mol} \quad 1 \mathrm{~mol} \\
& \frac{10}{2} \mathrm{~mol} \quad \frac{64}{32} \mathrm{~mol} \\
& =5 \mathrm{~mol}=2 \mathrm{~mol} \\
& \because \frac{1}{2} \mathrm{molO}_2 \text { gives }=1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} \\
& \therefore 2 \mathrm{~mol} \mathrm{O}_2 \text { will give }=1 \times 2 \times 2=4 \mathrm{~mol} .
\end{aligned}
$$
(i) $\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}$
(ii) Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount)
$$
\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 1 \mathrm{~mol} \quad \frac{1}{2} \mathrm{~mol} \quad 1 \mathrm{~mol} \\
& \frac{10}{2} \mathrm{~mol} \quad \frac{64}{32} \mathrm{~mol} \\
& =5 \mathrm{~mol}=2 \mathrm{~mol} \\
& \because \frac{1}{2} \mathrm{molO}_2 \text { gives }=1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} \\
& \therefore 2 \mathrm{~mol} \mathrm{O}_2 \text { will give }=1 \times 2 \times 2=4 \mathrm{~mol} .
\end{aligned}
$$
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