Question
What quantity of copper oxide will react with $$2.80\,L$$ of hydrogen at $$NTP?$$
A.
$$79.5\,g$$
B.
$$2\,g$$
C.
$$9.9\,g$$
D.
$$22.4\,g$$
Answer :
$$9.9\,g$$
Solution :
$$\eqalign{
& \mathop {CuO}\limits_{79.5\,g} + \mathop {{H_2}}\limits_{22.4\,L} \to Cu + {H_2}O \cr
& 22.4\,L\,\,{\text{of}}\,\,{H_2} \equiv 79.5\,g\,\,{\text{of}}\,\,CuO \cr} $$
$$2.80\,\,L\,\,{\text{of}}\,\,{H_2} \equiv \frac{{79.5}}{{22.4}} \times 2.80$$ $$ = 9.9\,g\,\,{\text{of}}\,\,CuO$$