Question

11.5 g of Na reacts with 9 g of $H_{2}O$. Percentage yield of the reaction is 80%. Calculate the number of molecules of $H_2$ produced in the reaction.

• A. 8.6$\times {10}^{22}$
• B. $12.05\times {10}^{22}$
• C. 8.6$\times {10}^{23}$
• D. 9.6$\times {10}^{23}$

Answer 1

The correct option is B $12.05\times {10}^{22}$

The reaction,
$Na+H_{2}O\to NaOH+\frac{1}{2}{H}_2$
Moles of Na present initially = $\frac{11.5}{23}=0.5moles$
From stoichiometry,
1 mole Na gives 0.5 moles of $H_2$
$\therefore$ 0.5 moles Na give 0.25 moles of $H_2$

But with 80% yield, moles of $H_2$ produced in the reaction = $0.25\times \frac{80}{100}=0.2moles$
So, mass of $H_2$ = $0.2\times 2=0.4g$
number of $H_2$ molecules = number of moles$\times$$N_A$, where $N_A$ is avogadro constant
So, number of $H_2$ molecules = $0.2\times 6.023\times {10}^{23}\simeq 12.05\times {10}^{22}molecules$