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$1 \mathrm{~mol}$ alcohol reacts with Na to give what weight of hydrogen?
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Verified Answer
The correct answer is:
$1 \mathrm{~g}$
$2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+2 \mathrm{Na} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2}$
$\mathrm{mol}$
mo
$\mathrm{mol}$
$$
\frac{1}{2} \mathrm{~mol}
$$
The weight of $\frac{1}{2} \mathrm{~mol} \mathrm{H}_{2}=\frac{2}{2} \mathrm{~g}=1 \mathrm{~g}$
Thus, $1 \mathrm{~g} \mathrm{H}_{2}$ is produced when 1 mole of alcohol reacts with Na.
$\mathrm{mol}$
mo
$\mathrm{mol}$
$$
\frac{1}{2} \mathrm{~mol}
$$
The weight of $\frac{1}{2} \mathrm{~mol} \mathrm{H}_{2}=\frac{2}{2} \mathrm{~g}=1 \mathrm{~g}$
Thus, $1 \mathrm{~g} \mathrm{H}_{2}$ is produced when 1 mole of alcohol reacts with Na.
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