\(n_{HCl}=0,2\cdot1,5=0,3\left(mol\right)\)
Đặt số mol của CuO, Fe2O3 lần lượt là a,b (mol)
PTHH:
\(CuO+2HCl->CuCl_2+H_2O\)
a --> 2a a a (mol)
\(Fe_2O_3+6HCl->2FeCl_3+3H_2O\)
b ---> 6b 2b 3b (mol)
Ta có pt
\(\left\{{}\begin{matrix}80a+160b=10\\2a+6b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,075\\b=0,025\end{matrix}\right.\)
\(\%m_{CuO}=\dfrac{m_{CuO}}{mhh}\cdot100\%=\dfrac{0,075\cdot80}{10}\cdot100\%=60\%\)
\(\%m_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{mhh}\cdot100\%=\dfrac{0,025\cdot160}{10}\cdot100\%=40\%\)
b) Pthh
\(CuO+CO\underrightarrow{t^o}Cu+CO_2\)
0,075-> 0,075 (mol)
\(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)
0,025--> 0,075 (mol)
\(V_{CO}=\left(0,075+0,075\right)\cdot22,4=3,36\left(l\right)\)
