B1: có $pV=nRT$ (R=0.082)
-> n= $\frac{pV}{RT}$
$n_{O_{2}}= \frac{1*24}{0.082*(27,3+273)} ≈ 0.975mol$
$n_{H_{2}} = \frac{2*15}{0.082*(25+273)} ≈1.228mol$
B2: $n_{CuSO_{4}.5H_{2}O}$ = $\frac{25}{250}$ =0.1mol
-> $C_{M}$ = $\frac{n_{CuSO_{4}.5H_{2}O}}{V_{dd}}$ =$\frac{0.1}{0.2}$ =0.5M
B3: $C$%= $\frac{m_{CuSO_{4}.5H_{2}O}}{m_{dd}}$ * $100$%= $12.5$%