$10 \mathrm{~g}$ of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water…
$10 \mathrm{~g}$ of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be
- $2 \mathrm{~mol}$
- $3 \mathrm{~mol}$
- $4 \mathrm{~mol}$
- $1 \mathrm{~mol}$
Solution
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)
$\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 1 \mathrm{~mol} \quad \frac{1}{2} \mathrm{~mol} \quad 1 \mathrm{~mol} \\
& \frac{10}{2} \mathrm{~mol} \quad \frac{64}{32} \mathrm{~mol} \\
& =5 \mathrm{~mol}=2 \mathrm{~mol} \\
& \because \frac{1}{2} \mathrm{molO}_2 \text { gives }=1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} \\
& \therefore 2 \mathrm{~mol} \mathrm{O}_2 \text { will give }=1 \times 2 \times 2=4 \mathrm{~mol} .
\end{aligned}$
Asked in: NEET 2009 (Screening)
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