Question

$23g$ of sodium will react with methyl alcohol to give:

• A. one mole of oxygen
• B. $22.4d{m}^3$ of hydrogen gas at NTP
• C. $1$ mole of $H_2$
• D. $11.2L$ of hydrogen gas at NTP

Answer 1

The correct option is D $11.2L$ of hydrogen gas at NTP

$Na+C{H}_{3}OH⟶C{H}_{3}ONa+\frac{1}{2}{H}_2$
Thus, $1$ $mole$ of $Na$ gives $\frac{1}{2}$ $mole$ of $H_2$.
In other words, $23$ $g$ of $Na$ gives $\frac{1}{2}\times 22.4=11.2$ $L$ of $H_2$ at NTP