Đáp án + Giải thích các bước giải:

Bài 3:

$n_{CuSO_4}=\frac{40.10\%}{160}=0,025(mol)$

$Zn+CuSO_4\to ZnSO_4+Cu$

Theo PT: $n_{Zn}=n_{ZnSO_4}=n_{Cu}=n_{CuSO_4}=0,025(mol)$

$\to m_{Zn}=0,025.65=1,625(g)$

Ta có $m_{ZnSO_4}=0,025.161=4,025(g);m_{Cu}=0,025.64=1,6(g)$

$\to C\%_{ZnSO_4}=\frac{4,025}{1,625+40-1,6}.100\%\approx 10,06\%$

Bài 4:

$a\big)Zn+H_2SO_4\to ZnSO_4+H_2$

$b\big)$

$n_{H_2}=\frac{1,12}{22,4}=0,05(mol)$

Theo PT: $n_{Zn}=n_{H_2}=0,05(mol)$

$\to m_{Zn}=0,05.65=3,25(g)$

$\to \%m_{Zn}=\frac{3,25}{5,25}.100\%\approx 61,9\%$

$\to \%m_{Cu}=100-61,9=38,1\%$

Oni