Question

$HNO_3$ oxidises $NH^{+}_4$ ions to nitrogen and itself gets reduced to $NO_2.$ The moles of $HNO_3$ required by $1\, mole$ of $\left(NH_4\right)2_SO_4$ is:

• A. $4$
• B. $5$
• C. $6$
• D. $2$

Answer 1

Answer: (C)

$HNO_3 + NH_4^{+} \to N_2 + NO_2$
$V.F. of HNO_3 = \left(5 - 4\right) = 1$
$V.F. of NH_4^{+} = \left[0 - (-3)\right] = 3$
so molar ratio of $HNO_3$ and $NH_4^{+} is 3 : 1.$
$1$ mole $\left(NH_4\right)_2SO_4$ is found to contain $2$ mole of $NH_4^{+}$