Đáp án:
a, $n_{CO2}$ = $\frac{m}{M}$ = $\frac{2}{12+16.2}$ = $\frac{1}{22}$ ≈ 0,045 mol
b, $n_{CuO}$ = $\frac{m}{M}$ = $\frac{16}{64+16}$ = 0,2 mol
c, $n_{H2}$ = $\frac{V}{22,4}$ = $\frac{11,2}{22,4}$ = 0,5 mol
d, $n_{O2}$ = $\frac{V}{22,4}$ = $\frac{5,6}{22,4}$ = 0,25 mol