a,
$m_{\text{giảm}}=13-12,75=0,25g=m_{Zn\text{pứ}}-m_{Cu}$
Gọi số mol Cu là x.
$Zn+CuSO_4\to ZnSO_4+Cu$
$\Rightarrow n_{Zn\text{pứ}}=n_{CuSO_4}=x (mol)$
$\Rightarrow 65x-64x=0,25$
$\Leftrightarrow x=0,25$
$\to m_{Zn}=0,25.65=16,25g$
b,
$C_{M_{CuSO_4}}=\dfrac{0,25}{0,2}=1,25M$