The quantity of copper (II) oxide reacts with 2.80 litre of hydrogen at NTP is

$\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$

$\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$

$1 \mathrm{mole} \mathrm{CuO}=1 \mathrm{mole} {\mathrm{H}}_2$

$79.5\mathrm{g} \mathrm{of} \mathrm{CuO}=22.4\mathrm{L} \mathrm{of} {\mathrm{H}}_2$

$?=2.8\mathrm{L} \mathrm{of} {\mathrm{H}}_2 \mathrm{at} \mathrm{NTP}$

$79.5\mathrm{g}=22.4\mathrm{L}$

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