Bạn tham khảo:
$Zn+H_2SO_4 \to ZnSO_4+H_2$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$n_{Zn}=n_{H_2}=0,1(mol)$
$a/$
$\%m_{Zn}=\dfrac{0,1.65}{13}.100\%=50\%$
$\%m_{Cu}=50\%$
$b/$
$n_{H_2SO_4}=n_{H_2}=0,1(mol)$
$C\%_{H_2SO_4}=\dfrac{0,1.98}{400}.100\%=2,45\%$