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Ammonia

HNO3 oxidises...

Question

$H N O_{3}$ oxidises $N H_{4}^{+}$ ions to nitrogen and itself get reduced to $N O_{2}$ . The mole of $H N O_{3}$ required by 1 mole of $(N H_{4})_{2} S O_{4}$ is

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Solution

$(N H_{4})_{2} S O_{4} + 6 H N O_{3} ⟶ N_{2} + 6 N O_{2}$
$1$ mole of $(N H_{4})_{2} S O_{4}$ uses $6$ moles of $H N O_{3}$ .

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