Balance the following equation.
$C u O + N H_{3} ⟶ C u + N_{2} + H_{2} O$

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Solution

Oxidation: $3 - N H_{3} ⟶ {0 N}_{2} + 3 e^{-} . . . . (1)$
Reduction: $2 + C u O + 2 e^{-} ⟶ 0 C u . . . . (2)$

Balancing $H$ atom in $(1)$
$N H_{3} ⟶ \frac{1}{2} N_{2} + 3 H^{+} + 3 e^{-} . . . . (3)$
Balancing $O$ & $H$ atoms in $(2)$ ,
$C u O + 2 e^{-} + 2 H^{+} ⟶ C u + H_{2} O . . . . (4)$
$(3) \times 2 + (4) \times 3$
$2 N H_{3} + 3 C u O ⟶ N_{2} + 3 C u + 3 H_{2} O$
This is the balanced equation.

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C u O + N H_{3} \to C u + N_{2} + H_{2} O

C u O + N H_{3} ⟶ C u + H_{2} O + N_{2}

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Balance the following equations :

(i) Na + $O_{2}$ $⟶$ $N a_{2} O$

(ii) $H_{2} O_{2}$ + $⟶$ $H_{2} O + O_{2}$

(iii) $M g (O H)_{2}$ +HCl $⟶$ $M g C l_{2}$ + $H_{2} O$

(iv) Fe + $O_{2}$ $⟶$ $F e_{2} O_{3}$

(v) $A l (O H)_{3}$ $⟶$ $A l_{2} O_{3}$ + $H_{2} O$

(vi) $N H_{3}$ + CuO $⟶$ Cu + $N_{2}$ + $H_{2} O$

(vii) $A l_{2} (S O_{4})_{3}$ + NaOH $⟶$ $A l (O H)_{3}$ + $N a_{2} S O_{4}$

(viii) $H N O_{3}$ + $C a (O H)_{2}$ $⟶$ $C a (N O_{3})_{2}$ + $H_{2} O$

(ix) NaOH + $H_{2} S O_{4}$ $⟶$ $N a_{2} S O_{4}$ + $H_{2} O$

(x) $B a C l_{2}$ + $H_{2} S O_{4}$ $⟶$ $B a_{2} S O_{4}$ + $H C l$

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