Bạn tham khảo
$a/$
$Mg+H_2SO_4 \to MgSO_4+H_2$
$n_{Mg}=0,25(mol)$
$n_{MgSO_4}=n_{H_2}=0,25(mol)$
$m_{MgSO_4}=0,25.120=30(g)$
$b/$
$V_{H_2}=0,25.22,4=5,6(l)
$c/$
$CuO+H_2 \xrightarrow{t^{o}} Cu+H_2O$
$n_{CuO}=0,3(mol)$
$H_2$ hết; $CuO$ dư
$n_{Cu}=0,25(mol)$
$m_{Cu}=0,25.64=16,(g)$
$m_{CuO(dư)}=(0,3-0,25)80=4(g)$