Making use of any of the substances given, write the balanced equations for the reactions by which you could obtain sulphur dioxide gas.
Sodium carbonate, sulphuric acid, sodium sulphite, lead, calcium carbonate.

2 N a_{2} {S O}_{3} + H_{2} {S O}_{4} (d i l) \to {N a}_{2} {S O}_{4} + H_{2} O + 2 {S O}_{2}

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N a_{2} {S O}_{3} + 2 H_{2} {S O}_{4} (d i l) \to {N a}_{2} {S O}_{4} + 2 H_{2} O + {S O}_{2}

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N a_{2} {S O}_{3} + H_{2} {S O}_{4} (d i l) \to {N a}_{2} {S O}_{4} + H_{2} O + {S O}_{2}

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2 N a_{2} {S O}_{3} + H_{2} {S O}_{4} (d i l) \to 2 {N a}_{2} {S O}_{4} + H_{2} O + {S O}_{2}

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Solution

The correct option is C $N a_{2} {S O}_{3} + H_{2} {S O}_{4} (d i l) \to {N a}_{2} {S O}_{4} + H_{2} O + {S O}_{2}$
By using sodium sulphite and sulphuric acid, sodium sulphate is prepared along with sulpher dioxide gas and water. $N a_{2} {S O}_{3} + H_{2} {S O}_{4} (d i l) \to {N a}_{2} {S O}_{4} + H_{2} O + {S O}_{2}$

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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

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

S + O_{2} \to {S O}_{2}; Δ H = - 298.2 k J

{S O}_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

{S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

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

S + O_{2} \to {S O}_{2}; △ H = - 298.2 k J

{S 0}_{2} + 1 / 20_{2} \to {S O}_{3}; △ H = - 98.7 k J

S O_{3} + H_{2} O \to H_{2} {S O}_{4}; △ H = - 130.2 k J

H_{2} + 1 / 20_{2} \to H_{2} O; △ H = - 287 k J

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

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