CHÚC BẠN HỌC TỐT!
Theo đề bài, ta có:\(\left\{{}\begin{matrix}n_{HCl}=0,7.2=1,4\left(mol\right)\\m_{ddHCl}=700.1,2=840\left(g\right)\end{matrix}\right.\)
Gọi x, y lần lượt là sô mol của CuO và Fe2O3
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
pư...........x............2x..............x..............x (mol)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
pư.............y...............6y................2y............3y (mol)
Ta có:\(\left\{{}\begin{matrix}m_{hh}=40\left(g\right)\\n_{HCl}=1,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}80x+160y=40\\2x+6y=1,4\end{matrix}\right.\)
Giải hệ suy ra:\(\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
a) \(C\%_{ddsaupư}=\dfrac{m_{ddsaupư}}{m_{hh}+m_{ddHCl}}.100\%\)
\(\Rightarrow C\%_{ddsaupư}=\dfrac{135.0,1+162,5.2.0,2}{40+840}.100\%\approx8,92\%\)
b) Ta có:\(\dfrac{m_{CuO}=80.0,1=8\left(g\right)}{m_{Fe2O3}=40-8=32\left(g\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}\%CuO=\dfrac{8}{40}.100\%=20\%\\\%Fe_2O_3=\dfrac{32}{40}.100\%=80\%\end{matrix}\right.\)
Vậy...............
