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Sulphuric Acid

The action of...

Question

The action of concentrated sulphuric acid on urea results in the production of a white crystalline solid X of formulae $H_{3} {N O}_{3} S$ , a monobasic acid. On treatment with sodium nitrite and dilute hydrochloric acid at $^{o} C$ , it liberates one mole of $N_{2}$ per mol of X, and on addition of aqueous barium chloride, the resulting solution yields one mole of ${B a S O}_{4}$ per mol of X taken. Here, X is $N H_{2} S O_{3} H$

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Solution

The action of concentrated sulphuric acid on urea results in the production of a white crystalline solid $N H_{2} S O_{3} H$ . It is a
monobasic acid.
$N H_{2} C O N H_{2} u r e a + H_{2} S O_{4} c o n c . ⟶ N H_{2} S O_{3} H X + N H_{3} + C O_{2}$
On treatment with sodium nitrite and dilute hydrochloric acid at $^{o} C$ , it liberates one mole of
$N_{2}$ per mole of $N H_{2} S O_{3} H$ .
$N a N O_{2} + H C l ⟶ H N O_{2} + N a C l$
$N H_{2} S O_{3} H + H N O_{2} ⟶ N_{2} ↑ + H_{2} S O_{4} + H_{2} O$ and on addition of aqueous barium
chloride, the resulting solution yields one mole of ${B a S O}_{4}$ per mol of $N H_{2} S O_{3} H$ taken. $H_{2} S O_{4} + B a C l_{2} ⟶ B a S O_{4} ↓ + 2 H C l$

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