$\left \{ {{Mg:a} \atop {Cu:b}} \right.$+$H_{2}$$SO_{4}$ →$\left \{ {{SO2:0,05mol} \atop {ddB}} \right.$
Bảo toàn e : 2a+2b=2.0,05=0,1 (I)
Dd B$\left \{ {{CuSO4} \atop {MgSO4}} \right.$ +NaOH→$\left \{ {{Cu(OH)2} \atop {Mg(OH)2}} \right.$ →$\left \{ {{MgO} \atop {CuO}} \right.$ +$H_{2}$ →$\left \{ {{MgO:a} \atop {Cu:b}} \right.$
⇒$m_{rắn}$ =40a+64b=2,72(g) (II)
(I)(II)⇒$\left \{ {{a=0,02} \atop {b=0,03}} \right.$ ⇒ Khối lượng
b)Dd B$\left \{ {{CuSO4:0,03} \atop {MgSO4:0,02}} \right.$ + H2O
Bảo toàn S:$n_{H2SO4}$= $n_{SO2}$ + $n_{SO4(muối)}$ =0,05+0,05=0,1 mol
⇒$m_{dd H2SO4}$= $\frac{0,1.98}{70}$.100 =14(g)
⇒$m_{dd B}$= $m_{A}$+ $m_{dd H2SO4}$-$m_{SO2}$ =13,2 (g) + 6,8 g=20(g)
⇒C%(CuSO4)=24%
C%(MgSO4)=12%