1 mole of methyl amine on reaction with nitrous acid gives at NTP

1.0 L of nitrogen

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22.4 L of nitrogen

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11.2 L of nitrogen

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5.6 L of nitrogen

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Solution

The correct option is D 22.4 L of nitrogen
${C H}_{3} - {N H}_{2} + {H N O}_{2} ⟶ {C H}_{3} O H + N_{2} + H_{2} O$
1 mole of methyl amine gives 1 mole $N_{2}$ ie, 22.4 L of nitrogen at NTP.

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