Question

When equal weights of methyl alcohol and ethyl alcohol react with excess of sodium metal the volume of $H_2$ liberated is more in the case of:

• A. $C_2{H}_{5}OH$
• B. $C{H}_{3}O$
• C. $Equalinboth$
• D. $H_2$ do not liberate

Answer 1

Answer: (A)

$C_2{H}_{5}OH$

$2C_2{H}_{5}OH+2Na⟶2C_2{H}_{5}ONa+H_2$

$2C{H}_{3}OH+2Na⟶2C{H}_{3}ONa+H_2$

Thus $2$ mole of Ethanol give $1$ mole of $H_2$

or $2$ mole of Methanol give $1$ mole of $H_2$

If $C{H}_{3}OH$ & $C_2{H}_{5}OH$ has equal weight, then number of moles of $C{H}_{3}OH=\frac{x}{32}$ mole

Number of moles of $C_2{H}_{5}OH=\frac{x}{46}$ mole

as $\frac{x}{32}>\frac{x}{46}$

Thus more number of moles of methanol will be there which gives more number of moles of $H_2$ gas or volume of $H_2$ gas.