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Q.

The volume of hydrogen required to reduce the 0.795g of CuO into Cu and $H_{2} O$ at STP is

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a

0.224 L

b

11.2 L

c

22.4 L

d

2.24 L

answer is B.

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Detailed Solution

$CuO + H_{2} \to Cu + H_{2} O$

$79.5 g of CuO \to 22.4 lit of H_{2}$

$0.795 g \to ?$

$= \frac{0.795}{79.5} \times 22.4$

= 0.224 lit.

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The volume of hydrogen required to reduce the 0.795g of CuO into Cu and H2O at STP is