Write balanced equation for reaction between sodium nitrate and concentrated sulphuric acid.

N a {N O}_{3} + 2 H_{2} {S O}_{4} \to 2 {N a}_{2} {S O}_{4} + {4 H N O}_{3}

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4 N a {N O}_{3} + H_{2} {S O}_{4} \to 2 {N a}_{2} {S O}_{4} + {2 H N O}_{3}

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2 N a {N O}_{3} + H_{2} {S O}_{4} \to {N a}_{2} {S O}_{4} + {2 H N O}_{3}

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N a {N O}_{3} + H_{2} {S O}_{4} \to {N a}_{2} {S O}_{4} + {2 H N O}_{3}

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Solution

The correct option is C $2 N a {N O}_{3} + H_{2} {S O}_{4} \to {N a}_{2} {S O}_{4} + {2 H N O}_{3}$
Sodium nitrate reacts with concentrated sulphuric acid to form sodium sulphate and nitric acid.

The balanced chemical reaction is $2 N a N O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H N O_{3}$ .

In the left-hand side and in the right-hand side of the chemical reaction the number of all atoms are the same. So the equation is perfectly balanced.

So the correct option is C.

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H_{2} {S O}_{4}

(a) H_{2} {S O}_{4} + N a O H \to N a H {S O}_{4} + H_{2} O

(b) H_{2} {S O}_{4} + 2 N a O H \to {N a}_{2} {S O}_{4} + 2 H_{2} O

K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

K M n O_{4} + {N a}_{2} {S O}_{3} \to M n O_{2} + {N a}_{2} {S O}_{4} + K O H

C u + H {N O}_{3} \to C u {({N O}_{3})}_{2} + {N O}_{2} + H_{2} O

K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

{N a}_{2} {S O}_{4}

N a O H

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

Identify the monoatomic and polyatomic ions from the following equation-
$P b (N O_{3})_{2}$ + $K_{2} S O_{4}$ $\to$ $2 K N O_{3}$ + $P b S O_{4}$

{2 S O}_{2} + O_{2} 450^{0} C - -- \to V_{2} O_{2} {2 S O}_{x} {S O}_{3} + H_{2} {S O}_{4} \to H_{2} S_{2} O_{y} H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} {S O}_{4}

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