$m_{chất \ rắn}=m_{Cu}=1,5(g)\\ n_{H_2}=0,2(mol)\\ Đặt\ n_{Fe}=x(mol)\\n_{Al}=y(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \to $$\left \{ {{56x+27y+1,5=7} \atop {x+1,5y=0,2}} \right.$ $\to$$\left \{ {{x=0,05} \atop {y=0,1}} \right.$
$\to m_{Fe}=0,05.56=2,8(g)\\m_{Al}=0,1.27=2,7(g)\\m_{Cu}=1,5(g)$
$c, n_{H_2SO_4}=0,05+0,1.1,5=0,2(mol)\\ m_{H_2SO_4}=0,2.98=19,6(g)$
