The quantity of copper (II) oxide reacts with 2.80 litre of hydrogen at NTP is $\text{CuO}+{\text{H}}_2\to \text{Cu}+{\text{H}}_2\text{O}$

1 mole CuO = 1 mole ${\text{H}}_2$ .

$79.5\text{g}\text{of}\text{CuO}=22.4\text{Lit}\text{of}{\text{H}}_{\text{2}}$

$?=2.8\text{L}\text{of}{\text{H}}_{\text{2}}\text{at}\text{NTP}$

$?\text{ ---- 2}\text{.8 lit}$

$79.5\text{g}=22.4\text{lit}$

$=\frac{79.5\times 28}{22.4}=9.9375\text{gm}$