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Stoichiometric Calculations

The answer to...

Question

The answer to each question of this set is a single digit integer, ranging from 0 to 9:
$H N O_{3}$ oxidises $N H_{4}^{+}$ ions to nitrogen and itself gets reduced to $N O_{2}$ . The moles of $H N O_{3}$ required by 1 mol of $N H_{4}^{+}$ is

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Solution

$(N H_{4})_{2} C O_{3} + 2 H N O_{3} \to C O_{2} + H_{2} O + 2 N H_{4} N O_{3}$ .
From the equatin we can conclude:
1 mole of $H N O_{3}$ is required by 1 mole of $N H_{4}^{+}$ .

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Similar questions

H N O_{3}

N H_{4}^{+}

N O_{2}

H N O_{3}

(N H_{4})_{2} S O_{4}

H N O_{3}

N H_{4}^{+}

N O_{2}

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1 m o l

(N H_{4})_{2} S O_{4}

O_{2}

H_{2} O_{2}

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