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Stoichiometry

Consider the ...

Question

Consider the following unbalanced reaction that is occurring in acidic solution:
$C r_{2} O_{7}^{2 -} + N O_{2}^{-} \to C r^{3 +} + N O_{3}^{-}$
When the reaction is balanced, what is the coefficient on the nitrate ion?

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is B $3$
Given : $C r_{2} O_{7}^{2 -} + N O_{2}^{-} ⟶ C r^{3 +} + N O_{3}^{-}$
Step 1: $C r_{2} O_{7}^{2 -} ⟶ C r^{3 +}$ and $H N O_{2} ⟶ N O_{3}^{-}$
$⟶ R e d u c t i o n$ $⟶ O x i d a t i o n$
Step 2:
$a .$ $C r_{2} O_{7}^{2 -} ⟶ C r^{3 +}$ and $H N O_{2} ⟶ N O_{3}^{-}$
$b .$ $C r_{2} O_{7}^{2 -} ⟶ C r^{3 +} + 7 H_{2} O$ and $H N O_{2} + H_{2} O ⟶ N O_{3}^{-}$
$c .$ $C r_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 2 C r^{3 +} + 7 H_{2} O$ & $H N O_{2} + H_{2} O ⟶ N O_{3}^{-} + 3 H^{+}$
$d .$ ${6 e}^{-} + {C r}_{2} O_{7}^{-} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 7 H_{2} O & H N O_{2} + H_{2} O ⟶ {N O}_{3}^{-} + {3 H}^{+} + {2 e}^{-}$
Step 3 : ${6 e}^{-} + {C r}_{2} O_{7}^{-} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 7 H_{2} O & 3 H N O_{2} + 3 H_{2} O ⟶ {2 N O}_{3}^{-} + {9 H}^{+} + {6 e}^{-}$
Adding both, ${C r}_{2} O_{7 (a g)}^{-} + 5 H_{(a g)}^{+} + 3 H N O_{2 (a g)} ⟶ 2 {C r}_{(a g)}^{3 +} + 3 N O_{3 (a g)}^{-} + 4 H_{2} O_{(l)}$
Coefficient on the nitrate ion is $3.$

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