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Sulphuric Acid

Sulphuric aci...

Question

Sulphuric acid is strong dehydrating agent. Give a chemical reaction in which the above property is exploited.

C_{6} H_{12} O_{6} \to 6 C + 6 H_{2} O

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

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H_{2} {S O}_{4} + P \to H_{3} {P O}_{4} + {S O}_{2} + H_{2} O

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

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Solution

The correct option is A $C_{6} H_{12} O_{6} \to 6 C + 6 H_{2} O$
Sulphuric acid reacts with glucose to produce carbon and hydrated sulphuric acid.
$6 H_{2} S O_{4}$ $+$ $C_{6} H_{12} O_{6}$ $⟶$ $6 C$ $+$ $6 (H_{2} S O_{4}$ $H_{2} O)$
Hence it removes the water from glucose.
(a) So this reaction shows the behaviour of Sulphuric acid as a strong dehydrating agent.
(b) Shows the decomposition of Sulphuric acid.
(d) Shows the acidic behaviour of Sulphuric acid.
(c) Shows the $H_{2} S O_{4}$ as oxidising agent.

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Similar questions

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}; △ 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}

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

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}

C u + H_{2} {S O}_{4} - \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

F e + + H_{2} O - \to {F e}_{3} O_{4} + H_{2} ↑

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

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