$3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 4 H_{2} O + 2 N O$ .
The volume of nitric oxide collected at the same time is :
[Cu=64, H=1, O=16, N=14]

5.6 L

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11.2 L

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22.4 L

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44.8 L

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Solution

The correct option is A 5.6 L
$3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 4 H_{2} O + 2 N O$
8 volume of nitric acid produce 2 volume of nitric oxide.
Volume of copper=22.4 L
22.4 L of nitric acid will produce $\frac{2}{4} \times 22.4 = 5.6 L$ .

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