Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\\n_{Fe_2O_3}=y\end{matrix}\right.\)
\(n_{HCl}=0,5.2=1mol\)
\(CuO+2HCl\rightarrow CuCl_2+H_2\)
x 2x ( mol )
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
y 6y ( mol )
Ta có:
\(\left\{{}\begin{matrix}80x+160y=32\\2x+6y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{CuO}=0,2.80=16g\)
\(\Rightarrow m_{Fe_2O_3}=0,1.160=16g\)
Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_2O_3}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
CuO + HCl ---> CuCl2 + H2O
a 2a
Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
b 6b
\(n_{HCl}=2.0,5=1\left(mol\right)\)
Hệ phương trình \(\left\{{}\begin{matrix}80a+160b=32\\2a+6b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)
