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What amount of oxygen is used at S.T.P to obtain $9 \mathrm{~g}$ water from sufficient amount of hydrogen gas?
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The correct answer is:
$5.6 \mathrm{dm}^3$
$$
\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\ell)}
$$
$11.2 \mathrm{dm}^3$ of $\mathrm{O}_2$ gives $18 \mathrm{~g}$ water at STP
$\therefore 9 \mathrm{~g}$ water is obtained from $\frac{112 \times 9}{18}=5.6 \mathrm{dm}^3$ of $\mathrm{O}_2$
\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\ell)}
$$
$11.2 \mathrm{dm}^3$ of $\mathrm{O}_2$ gives $18 \mathrm{~g}$ water at STP
$\therefore 9 \mathrm{~g}$ water is obtained from $\frac{112 \times 9}{18}=5.6 \mathrm{dm}^3$ of $\mathrm{O}_2$
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