Question

The volume of methane at NTP formed from $8.2g$ of sodium acetate by fusion with soda lime is:

• A. $10lit$
• B. $11.2lit$
• C. $5.6lit$
• D. $3.75lit$

Answer 1

The correct option is D $3.75lit$

The observation reaction can be written as-

$C{H}_{3}COONa+NaOH⟶C{H}_4+N{a}_{2}C{O}_3$

From the above reaction, it is cleared that $1mole$ of $C{H}_{3}COONa$

will produce $1mole$ of $C{H}_4$

Molecular weight of sodium Acetate $\left(C{H}_{2}COONa\right)=82g$

No. of moles in $8.2g$ of $C{H}_{3}COONa=\frac{8.2}{82}$ $=0.1mole$

Hence, $0.1mole$ of $C{H}_{3}COONa$ will produce $0.1mole$ of $C{H}_4$.

Molecular weight of $C{H}_4$ is $16$

$0.1mole$ of methane $=16\times 0.1=1.6g$ of methane

Density of methane $=0.4256g/ml$

Volume = Mass / Density

$=\frac{1.6}{0.4256}=3.75ml$

Hence the volume of methane $\left(C{H}_4\right)$ produced will be $3.75ml$

So, the correct option is $D$