Đáp án + Giải thích các bước giải:

$a\bigg)$

$n_{H_2}=\frac{3,136}{22,4}=0,14(mol)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$FeO+H_2\xrightarrow{t^o}Fe+H_2O$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O$

Theo PT: $n_{H_2O}=n_{H_2}=0,14(mol)$

Bảo toàn khối lượng: 

$m_A+m_{H_2}=m_B+m_{H_2O}$

$\to m_B=8,48+0,14.2-0,14.18=6,24(g)$

$b\bigg)$

Đặt $n_{FeO}=2x(mol)\to n_{CuO}=x(mol)$

Đặt $n_{Fe_2O_3}=y(mol)\to n_{Fe_3O_4}=2y(mol)$

Ta có $m_{FeO}+m_{CuO}+m_{Fe_2O_3}+m_{Fe_3O_4}=8,48$

$\to 2x.72+80x+160.y+232.2y=8,48$

$\to 224x+624y=8,48(1)$

Theo PT: $n_{CuO}+n_{FeO}+3n_{Fe_2O_3}+4n_{Fe_3O_4}=n_{H_2}$

$\to x+2x+3y+4.2y=0,14$

$\to 3x+11y=0,14(2)$

Từ $(1)(2)\to x=0,01(mol);y=0,01(mol)$

$\to \begin{cases} m_{FeO}=0,01.2.72=1,44(g)\\ m_{CuO}=0,01.80=0,8(g)\\ m_{Fe_2O_3}=0,01.160=1,6(g)\\ m_{Fe_3O_4}=0,01.2.232=4,64(g) \end{cases}$

$\%m_{FeO}=\frac{1,44}{8,48}.100\%\approx 16,98\%$

$\%m_{CuO}=\frac{0,8}{8,48}.100\%\approx 9,43\%$

$\%m_{Fe_2O_3}=\frac{1,6}{8,48}.100\%\approx 18,87\%$

$\%m_{Fe_3O_4}=100-18,87-9,43-16,98=54,72\%$