$10 \mathrm{~g}$ of hydrogen and $64 \mathrm{~g}$ of oxygen were filled in a steel vessel and exploded.…
$10 \mathrm{~g}$ of hydrogen and $64 \mathrm{~g}$ of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be -
$1 \mathrm{~mol}$
$2 \mathrm{~mol}$
$3 \mathrm{~mol}$
$4 \mathrm{~mol}$
Solution
$\mathrm{H}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}$
$2 \mathrm{~g} \quad 16 \mathrm{~g} \quad 18 \mathrm{~g}$
$10 \mathrm{~g} \mathrm{H}_2$ required $\mathrm{O}_2=80$ which is not present 64 g $\mathrm{O}_2$ required $8 \mathrm{~g}$ of $\mathrm{H}_2$ and $\mathrm{H}_2$ left $=2 \mathrm{~g}$. Thus, $\mathrm{O}_2$ is the limiting reactant and $\mathrm{H}_2$ is excess reactant.
Hence, $\mathrm{H}_2 \mathrm{O}$ formed from 64 of $\mathrm{O}_2$
$\begin{aligned}
& =\frac{18}{16} \times 64 \\
& =72 \mathrm{~g}=\frac{72}{18} \mathrm{~mole} \\
& =4 \mathrm{~mole}
\end{aligned}$