a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\) (2)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\) (3)
b, Ta có: \(n_{H_2}=0,3\left(mol\right)\)
Theo PT (1): \(\left\{{}\begin{matrix}n_{Al}=\frac{2}{3}n_{H_2}=0,2\left(mol\right)\\n_{HCl\left(1\right)}=2n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{FeO}+m_{Fe_2O_3}=35,8-m_{Al}=35,8-0,2.27=30,4\left(g\right)\)
Ta có: \(\Sigma n_{HCl}=1,6.1=1,6\left(mol\right)\)
\(\Rightarrow n_{HCl\left(2\right)}+n_{HCl\left(3\right)}=1,6-n_{HCl\left(1\right)}=1,6-0,6=1\left(mol\right)\)
Giả sử: \(\left\{{}\begin{matrix}n_{FeO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow72x+160y=30,4\left(1\right)\)
Theo PT (2) + (3): \(\Sigma n_{HCl\left(2\right)+\left(3\right)}=2n_{FeO}+6n_{Fe_2O_3}=2x+6y\left(mol\right)\)
\(\Rightarrow2x+6y=1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%m_{FeO}=\frac{0,2.72}{35,8}.100\%\approx30,22\%\)
c, Muối thu được sau pư gồm: AlCl3; FeCl2 và FeCl3
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{FeO}=0,2\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{muoi}=m_{AlCl_3}+m_{FeCl_2}+m_{FeCl_3}\)
= 0,2.133,5 + 0,2.127 + 0,2.162,5
= 84,6 (g)
Bạn tham khảo nhé!
