Đáp án:
$V=8,96\\ m_{Cu}=1(g)\\ m_{Al}=2,7(g)\\ m_{Fe}=14(g)$
Giải thích các bước giải:
$m_{Cu}=1(g)\\ n_{H_2SO_4}=\dfrac{100.39,2}{98.100}=0,4(mol)\\ Đặt\ \begin{cases}n_{Fe}=x(mol)\\n_{Al}=y(mol)\end{cases}\\ \Rightarrow 56x+27y=17,7-1=16,7\ (1)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ n_{H_2}=n_{H_2SO_4}=0,4(mol)\\ V_{H_2}=0,4.22,4=8,96(l)\\ n_{H_2SO_4}=n_{Fe}+\dfrac{3}{2}.n_{Al}=x+\dfrac{3}{2}y(mol)\\ \Rightarrow x+\dfrac{3}{2}y=0,4\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,25\\y=0,1\end{cases}\\ m_{Fe}=0,25.56=14(g)\\ m_{Al}=0,1.27=2,7(g)$