Number of moles of $A s_{2} S_{3}$ required to reduce $56$ moles of $H N O_{3}$ according to equation,
$A s_{2} S_{3} + H N O_{3} ⟶ H_{3} A s O_{4} + H_{2} S O_{4} + N O$ , is:

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Solution

Balancing the reaction, we get the following equation:
$3 A s_{2} S_{3} + 28 H N O_{3} + 4 H_{2} O ⟶ 6 H_{3} A s O_{4} + 9 H_{2} S O_{4} + 28 N O$
3 28
6 56
As per the reaction, for $28$ moles of $H N O_{3}$ , $3$ moles of $A s_{2} S_{3}$ is required.
So, for $56$ moles of $H N O_{3}$ , $6$ moles of $A s_{2} S_{3}$ is required.

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