a,
$Zn+H_2SO_4\to ZnSO_4+H_2$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
Theo PTHH:
$n_{Zn}=n_{H_2SO_4}=n_{ZnSO_4}=0,1(mol)$
$\to m_{Zn}=0,1.65=6,5g$
Chất rắn là $Cu$
$\to m_{Cu}=10,5-6,5=4g$
b,
$V_{H_2SO_4}=\dfrac{0,1}{0,5}=0,2l$
$m_{ZnSO_4}=0,1.161=16,1g$