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What quantity of copper oxide will react with 2.80 L of hydrogen at NTP?

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79.5 g

2 g

9.9 g

22.4 g

answer is C.

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Detailed Solution

$\underset{79.5 g}{CuO} + \underset{22.4 L}{H_{2}} \to Cu + H_{2} O$
22.4 L of H₂ $\equiv$ 79.5 g of CuO
2.80 L of H₂ $= \frac{79 .5}{22 .4} \times 2 .80 = 9 .9 g of CuO$

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