Question

The volume (ml) of 3M $HN{O}_3$ required to oxidised 8 g $F{e}^{2+}$ is:
$(HN{O}_3+F{e}^{2+}\to F{e}^{3+}+NO+H_{2}O)$:

• A. 15.88
• B. 32.78
• C. 14.22
• D. 12.32

Answer 1

Answer: (A)

15.88

Number of moles in $8g$ of $F{e}^{+2}=\frac{8}{56}$ moles

Now, by

$\left(HN{O}_3\right)n_1\times n{f}_1=n_2\times n{f}_2\left(F{e}^{+2}\right)$

$\Rightarrow n_1\times 3=\frac{8}{56}$ moles of $F{e}^{+2}$ as $N^{+5}⟶N^{+2},nf$ for $HN{O}_3=3$

Thus $n_1$ mole of $HN{O}_3$ are required to oxidize $8g$ of $F{e}^{+2}$

$\Rightarrow n_1=M_1{V}_1=3\times V_1$

$\Rightarrow 3\times 3\times V_1=\frac{8}{56}$ moles

$\Rightarrow V_1=\frac{8\times 1000}{56\times 3}=\frac{47.61}{3}ml=15.88ml$

Thus, $\frac{47.61}{3}ml$ of $HN{O}_3$ is required to oxidise $8g$ of $F{e}^{+2}$ .