Question

$1.15\text{g}$ of $Na$ reacts with $0.9\text{g}$ of $H_{2}O$. The percentage yield of the reaction is $80\%$. Calculate the number of molecules of $H_2$ produced in the reaction.

• A. $8.6\times {10}^{20}$
• B. $12.05\times {10}^{21}$
• C. $8.6\times {10}^{21}$
• D. $9.6\times {10}^{21}$

Answer 1

The correct option is B $12.05\times {10}^{21}$

The reaction,
$Na+H_{2}O\to NaOH+\frac{1}{2}{H}_2$
Moles of $Na$ present initially $=\frac{1.15}{23}=0.05\text{mol}$
From stoichiometry,
$1$ mole of $Na$ gives $0.5$ moles of $H_2$
$\therefore$ $0.05$ moles of $Na$ give $0.025$ moles of $H_2$

But with $80\%$ yield, moles of $H_2$ produced in the reaction $=0.025\times \frac{80}{100}=0.02\text{mol}$
So, mass of $H_2$ $=0.02\times 2=0.04\text{g}$
Number of $H_2$ molecules $=$ number of moles$\times$$N_A$, where $N_A$ is Avogadro constant.
So, number of $H_2$ molecules $=0.02\times 6.023\times {10}^{23}\simeq 12.05\times {10}^{21}\text{molecules}$