Question

In acid medium $Zn$ reduces nitrate ion to $N{H}_4^{+}$ ion according to the reaction $Zn+N{O}_3^{-}\to Z{n}^{2+}$ $+N{H}_4^{+}+H_{2}O$ (unbalanced) How many moles of $HCl$ are required to reduce half a mole of $NaN{O}_3$ completely? Assume the availability of sufficient $Zn$

• A. 5
• B. 4
• C. 3
• D. 2
• E. 1

Answer 1

Answer: (A)

First the given unbalanced equation is balanced by using following steps Step I. The equation is splitted into two half equations as
$Zn\to Z{n}^{2+};$
$N{O}_3^{-}\to N{H}_4^{+}$
Step II. Now water molecules are added to the side deficient in oxygen and $H^{+}$
are added to the side deficient in hydrogen as
$Zn\to Z{n}^{2+};N{O}_3^{-}+10H^{+}\to N{H}_4^{+}+3H_{2}O$
Step III. The number of electrons are balanced and the two half equations are added. $[Zn\to Z{n}^{2+}+2e^{-}]\times 4;$
$N{O}_3^{-}+10H^{+}+8e^{-}\to N{H}_4^{+}+3H_{2}O$
$4Zn\to 4Z{n}^{2+}+8e^{-}$
$\therefore$ $4Zn+N{O}_3^{-}+10H^{+}\to 4Z{n}^{2+}+N{H}_4^{+}$
$+3H_{2}O$ (Net equation) or $4Zn+N{o}_3^{-}+10HCl\to 4Z{n}^{2+}+N{H}_4^{+}$
$+5C{l}_2+3H_{2}O$
$\because$ 1 mole of $N{O}_3^{-}$ (or $NaN{O}_3$ )is reduced by = 10 moles of $HCl$
$\therefore$ $\frac{1}{2}$ mole of $N{O}_3^{-}$ will be reduced by
$=10\times \frac{1}{2}molesofHCl$ = 5 moles of $HCl$