$3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 4 H_{2} O + 2 N O$ .
The mass of copper needed to react with 63 g of nitric acid is :

63 g

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12 g

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44 g

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24 g

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Solution

The correct option is A 24 g
$3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 4 H_{2} O + 2 N O$
Moles of nitric acid $= \frac{63}{63} = 1$
3 moles of copper reacts with 8 moles of nitric acid therefore 1 mole of nitric acid reacts with three-eighth mole of copper.
Therefore, mass of copper $= \frac{3}{8} \times 63.5$ $= 24$ g

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H N O_{3}

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