Balanced molecular equation for the laboratory preparation of nitric acid from $N a N O_{3}$ :

3 {N a N O}_{3} + 2 H_{2} {S O}_{4} (c o n c) \to {N a H S O}_{4} + 3 {H N O}_{3}

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{N a N O}_{3} + 4 H_{2} {S O}_{4} (c o n c) \to 4 {N a H S O}_{4} + {H N O}_{3}

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{N a N O}_{3} + H_{2} {S O}_{4} (c o n c) \to {N a H S O}_{4} + {H N O}_{3}

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2 {N a N O}_{3} + H_{2} {S O}_{4} (c o n c) \to 2 {N a H S O}_{4} + {H N O}_{3}

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Solution

The correct option is B ${N a N O}_{3} + H_{2} {S O}_{4} (c o n c) \to {N a H S O}_{4} + {H N O}_{3}$
Sodium nitrate reacts with conc. sulfuric acid to form sodium bisulfate and nitric acid.

The balanced equation is $N a N O_{3} + H_{2} S O_{4} (c o n c) \to N a H S O_{4} + H N O_{3}$ .

The reaction takes place at around $200^{\circ}$ C.

So the correct option is C.

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