Bạn tham khảo:
$a/$
$\%m_{Cu}=\frac{0,64}{2,72}.100\%=23,52\%$
$Mg+H_2SO_4 \to MgH_2SO_4+H_2$
$Fe+H_2SO_4 \to FeSO_4+H_2$
$m_{hh(Fe; Mg)}=2,72-0,64=2,08(g)$
$n_{Fe}=a(mol)$
$n_{Mg}=b(mol)$
$m_{hh}=56a+24b=2,08(1)$
$n_{H_2}=0,07(mol)$
$n_{H_2}=a+b=0,07(2)$
$(1)(2)$
$a=0,0125$
$b=0,0575$
$\%m_{Fe}=\frac{0,0125.56}{2,72}.100\%=25,7\%$
$\%m_{Mg}=50,78\%$
$b/$
$n_{FeSO_4}=0,0125(mol)$
$n_{MgSO_4}=0,0575(mol)$
$m_{muối}=0,0125.152+0,0575.120=8,8(g)$