`n_(H_2)=\frac{2,24}{22,4}=0,1(mol)`

`Fe+H_2SO_4->FeSO_4+H_2`

`0,1`      `0,1`       `0,1`       `0,1`

`Al_2O_3+3H_2SO_4>Al_2(SO_4)_3+3H_2O`

`0,2`              `0,6`          `0,2` 

`m_(Fe)=0,1.56=5,6(g)`

`=>m_(Al_2O_3)=26-5,6=20,4(g)`

`n_(Al_2O_3)=\frac{20,4}{102}=0,2(mol)`

`m_(H_2SO_4)=0,7.98=68,6(g)`

Lại có axit dư `10%`

`=>m_(HCl  dư)=\frac{68,6}{90%}-68,6=\frac{343}{45}(g)`

`m_(dd)=26+200-0,2=225,8(g)`

`C%_(FeSO_4)=\frac{0,1.152}{225,8}.100=6,73%`

`C%_(Al_2(SO_4)_3)=\frac{0,2.342}{225,8}.100=30,3%`

`C%_(H_2SO_4)=\frac{\frac{343}{45}}{225,8}.100=3,376%`

Đáp án:

\(\begin{array}{l} C\%_{Al_2(SO_4)_3}=30,29\%\\ C\%_{FeSO_4}=6,73\%\\ C\%_{H_2SO_4}\text{(dư)}=3,37\%\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} PTHH:\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\ (1)\\ Fe+H_2SO_4\to FeSO_4+H_2↑\ (2)\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt\ (2):\ n_{Fe}=n_{H_2}=0,1\ mol.\\ \Rightarrow m_{Fe}=0,1\times 56=5,6\ g.\\ \Rightarrow m_{Al_2O_3}=26-5,6=20,4\ g.\\ \Rightarrow n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\ mol.\\ \Rightarrow ∑n_{H_2SO_4}=3n_{Al}+n_{Fe}=0,6+0,1=0,7\ mol.\\ \Rightarrow m_{H_2SO_4}=0,7\times 98=68,6\ g.\\ \Rightarrow m_{H_2SO_4}\text{(dư)}=\dfrac{68,6}{100\%-10\%}-68,6=7,62\ g.\\ m_{\text{dd spư}}=m_{\text{hỗn hợp}}+m_{\text{dd H$_2$SO$_4$}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=26+200-0,1\times 2=225,8\ g.\\ Theo\ pt\ (1):\ n_{Al_2(SO_4)_3}=n_{Al_2O_3}=0,2\ mol.\\ Theo\ pt\ (2):\ n_{FeSO_4}=n_{Fe}=0,1\ mol.\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2\times 342}{225,8}\times 100\%=30,29\%\\ C\%_{FeSO_4}=\dfrac{0,1\times 152}{225,8}\times 100\%=6,73\%\\ C\%_{H_2SO_4}\text{(dư)}=\dfrac{7,62}{225,8}\times 100\%=3,37\%\end{array}\)

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