B1: có p*V=n*R*T (R=0.082)
-> n= $\frac{pV}{RT}$
$n_{O{2}}$= $\frac{1*24}{0.082*(27,3+273)}$ ≈ 0.975mol
$n_{H{2}}$≈1.228mol
B2: $n_{CuSO4.5H2O}$ = $\frac{25}{250}$ =0.1mol
-> $C_{M}$ = $\frac{n_{CuSO4.5H2O}}{V_{dd}}$ =$\frac{0.1}{0.2}$ =0.5M
B3: C%= $\frac{m_{CuSO4.5H2O}}{m_{dd}}$ * 100%= 12.5%