Question

In acid medium $Zn$ reduces nitrate ion to $ NH_{4}^{+} $ ion according to the reaction $ Zn+NO_{3}^{-}\to Z{{n}^{2+}} $ $ +NH_{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\xrightarrow[{}]{{}}Z{{n}^{2+}}; $
$ NO_{3}^{-}\xrightarrow{{}}NH_{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\xrightarrow{{}}Z{{n}^{2+}};NO_{3}^{-}+10{{H}^{+}}\xrightarrow{{}}NH_{4}^{+}+3{{H}_{2}}O $
Step III. The number of electrons are balanced and the two half equations are added. $ [Zn\xrightarrow{\,}\,Z{{n}^{2+}}\,+2{{e}^{-}}]\,\times 4; $
$ NO_{3}^{-}+10{{H}^{+}}+8{{e}^{-}}\xrightarrow{{}}NH_{4}^{+}+3{{H}_{2}}O $
$ 4Zn\xrightarrow{{}}4Z{{n}^{2+}}+8{{e}^{-}} $
$ \therefore $ $ 4Zn+NO_{3}^{-}+10{{H}^{+}}\xrightarrow{{}}4Z{{n}^{2+}}+NH_{4}^{+} $
$ +3{{H}_{2}}O $ (Net equation) or $ 4Zn+No_{3}^{-}+10HCl\xrightarrow{{}}4Z{{n}^{2+}}+NH_{4}^{+} $
$ +5C{{l}_{2}}+3{{H}_{2}}O $
$ \because $ 1 mole of $ NO_{3}^{-} $ (or $ NaN{{O}_{3}} $ )is reduced by = 10 moles of $ HCl $
$ \therefore $ $ \frac{1}{2} $ mole of $ NO_{3}^{-} $ will be reduced by
$ =10\times \frac{1}{2}moles\text{ }of\text{ }HCl $ = 5 moles of $ HCl $