a) \(n_{HCl}=0,2.1,5=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) => 2x + y = 10 (*)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
x----->2x
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
y-------->6y
=> 2x + 6y = 0,3 (**)
Từ (*), (**) => \(\left\{{}\begin{matrix}x=0,075\\y=0,025\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,075.80}{10}.100\%=60\%\\\%m_{Fe_2O_3}=100\%-60\%=40\%\end{matrix}\right.\)
b) PTHH: \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\)
0,075->0,075
\(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\)
0,025--->0,075
=> \(V_{CO_2}=\left(0,075+0,075\right).22,4=3,36\left(l\right)\)
