\(n_{H_2}=\dfrac{11.2}{22,4}=0,5\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{Fe_3O_4}=b\left(mol\right)\end{matrix}\right.\)

\(160a+232b=27,2\left(g\right)\left(1\right)\) 

PTHH : Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O 

          \(a\)         \(3a\)             \(2a\)

PTHH : Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

         \(b\)         \(4b\)              \(3b\)           

\(3a+4b=0,5\left(mol\right)\) \(\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=\dfrac{9}{70}\left(mol\right)\\b=\dfrac{1}{35}\left(mol\right)\end{matrix}\right.\)

\(n_{Fe\left(2\right)}=3b=\dfrac{1}{35}.3=\dfrac{3}{35}\left(mol\right)\)

\(m_{Fe\left(2\right)}=\dfrac{3}{35}.56=4,8\left(g\right)\)

\(n_{Fe\left(1\right)}=2a=2.\dfrac{9}{70}=\dfrac{9}{35}\left(mol\right)\)

\(m_{Fe\left(1\right)}=\dfrac{9}{35}.56=14,4\left(g\right)\)

\(m_{Fe\left(tổng\right)}=4,8+14,4=19,2\left(g\right)\)

\(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)

\(3Ba\left(OH\right)_2+2AlCl_3->2Al\left(OH\right)_3+3BaCl_2\)

\(3Ca\left(OH\right)_2+P_2O_5->Ca_3\left(PO_4\right)_2+3H_2O\)

6. \(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)

7. \(3Ba\left(OH\right)_2+2AlCl_3->2Al\left(OH\right)_3+3BaCl_2\)

8. \(3Ca\left(OH\right)_2+P_2O_5->Ca_3\left(PO_4\right)_2+3H_2O\)

35 \(2C_2H_6O+2Na->2C_2H_5ONa+H_2\)

36. \(2C_4H_{10}+13O_2->8CO_2+10H_2O\)

a)\(m_{H_2SO_4}=\dfrac{m_{ddH_2SO_4}\cdot C\%}{100\%}=\dfrac{15\cdot10\%}{100\%}=1,5g\)

b)\(m_{NaHCO_3}=\dfrac{m_{ddNaHCO_3}\cdot C\%}{100\%}=\dfrac{34\cdot25\%}{100\%}=8,5G\)