\(n_{CuO}=x;n_{Fe_2O_3}=y\)
\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(PTHH:Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(hpt:\left\{{}\begin{matrix}80x+160y=3,2\\\frac{x}{2y}=\frac{1}{1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\\y=0,01\end{matrix}\right.\)
\(m_{CuO}=80x=80.0,02=1,6\left(g\right)\\ m_{Fe_2O_3}=160.0,01=1,6\left(g\right)\)
Đặt :
nCuO = x mol
nFe2O3 = y mol
<=> 80x + 160y = 3.2 (1)
CuO + 2HCl --> CuCl2 + H2O
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
<=> x = 2y
<=> x - 2y = 0 (2)
Giải (1) và (2) :
x = 0.02
y = 0.01
mCuO = 1.6 g
mFe2O3 = 1.6 g
%CuO = %Fe2O3 = 50%
