Balance the following equation.
$C u S O_{4} + N a O H ⟶ C u (O H)_{2} + N a_{2} S O_{4}$

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Solution

Here, number of $N a, O$ and $H$ atoms is different in $L H S$ & $R H S$ of the given equation.
We need to take $2$ moles of $N a O H$ to balance those atoms.
Hence, the balanced equation is:
$C u S O_{4} + 2 N a O H ⟶ C u (O H)_{2} + N a_{2} S O_{4}$

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Balance the following equation

Al₂(SO₄)₃+NaOH - Al(OH)₃+Na₂SO₄

N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O

NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + H_{2} O

N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O

What type of reactions are represented by the following equations

(a) $C a O + C O_{2} ⟶ C a C O_{3}$

(b) $2 N a + 2 H_{2} O ⟶ 2 N a O H + H_{2}$

(d) $N H_{4} N O_{2} ⟶ N_{2} + 2 H_{2} O$

(e) $C u S O_{4} + 2 N a O H ⟶ C u (O H)_{2} + N a_{2} S O_{4}$

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