a)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
b)
- Phần 1: Gọi \(\left(n_{Al_2O_3};n_{Fe_2O_3};n_{CuO}\right)=\left(a;b;c\right)\)
=> 102a + 160b + 80c = 4,22 (1)
nHCl = 0,2.0,8 = 0,16 (mol)
Theo PTHH: \(n_{HCl}=6a+6b+2c=0,16\) (2)
- Phần 2: Gọi \(\left(n_{Al_2O_3};n_{Fe_2O_3};n_{CuO}\right)=\left(ak;bk;ck\right)\)
=> ak + bk + ck = 0,08 (*)
Theo PTHH: \(n_{H_2O}=3bk+ck=\dfrac{1,8}{18}=0,1\left(mol\right)\) (**)
(*)(**) => \(\dfrac{a+b+c}{3b+c}=\dfrac{4}{5}\)
=> 5a - 7b + c = 0 (3)
(1)(2)(3) => a = 0,01 (mol); b = 0,01 (mol); c = 0,02 (mol)
\(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,01.102}{4,22}.100\%=24,17\%\\\%m_{Fe_2O_3}=\dfrac{0,01.160}{4,22}.100\%=37,91\%\\\%m_{CuO}=100\%-24,17\%-37,91\%=37,92\%\end{matrix}\right.\)
