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(A) The general reaction of an alcohol $(R-OH)$ with sodium $(Na)$ is given by:$2 R-OH + 2 Na \longrightarrow 2 R-ONa + H_2$From the balanced chemical

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(A) The general reaction of an alcohol $(R-OH)$ with sodium $(Na)$ is given by:$2 R-OH + 2 Na \longrightarrow 2 R-ONa + H_2$From the balanced chemical equation,$2 \ mol$ of alcohol produces $1 \ mol$ of hydrogen gas $(H_2)$.Therefore,$1 \ mol$ of alcohol produces $\frac{1}{2} \ mol$ of $H_2$.The molar mass of $H_2$ is $2 \ g/mol$.Weight of $H_2 = 	ext{moles} 	imes 	ext{molar mass} = \frac{1}{2} \ mol 	imes 2 \ g/mol = 1 \ g$.Thus,$1 \ g$ of hydrogen is produced.

$1 \ mol$ of alcohol reacts with $Na$ to produce what weight of hydrogen (in $g$)?

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