Ta có:
\(\frac{m_{Fe2O3}}{m_{CuO}}=\frac{2}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}3m_{Fe2O3}-2m_{CuO}=0\\480m_{Fe2O3}-160n_{CuO}=0\\160n_{Fe2O3}+80n_{CuO}=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Fe2O3}=0,08\left(mol\right)\\n_{CuO}=0,24\left(mol\right)\end{matrix}\right.\)
PTHH:
\(Fe_2O_3+3H_2\rightarrow3Fe+3H_2O\)
0,08____0,24______0,16__________
\(CuO+H_2\rightarrow Cu+H_2O\)
0,24___0,24____0,24_____
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,16.56}{0,16.56+0,24.64}.100\%=36,84\%\\\%m_{Cu}=100\%-36,84\%=63,16\%\end{matrix}\right.\)
\(\Rightarrow V_{H2}=0,48.22,4=10,752\left(l\right)\)
Đặt \(n_{Fe_2O_3}=x\left(mol\right);n_{CuO}=y\left(mol\right)\)
Theo đề bài ta có:
\(\left\{{}\begin{matrix}160x+80y=32\\\frac{x}{y}=\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{35}\\y=\frac{6}{35}\end{matrix}\right.\)
\(\%m_{Fe_2O_3}=\frac{\frac{160.4}{35}}{32}.100\%=57,1\left(\%\right)\Rightarrow\%m_{CuO}=100-57,1=42,9\left(\%\right)\)
\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
(mol)_____\(\frac{4}{35}\)______\(\frac{12}{35}\)____________
\(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
(mol)____\(\frac{6}{35}\)_____\(\frac{6}{35}\)__________
\(V_{H_2}=22,4.\left(\frac{6}{35}+\frac{12}{35}\right)=11,52\left(l\right)\)
Tính lại cái %m nhé, tại chưa đọc kĩ đề :)
\(\%m_{Fe}=\frac{\frac{56.8}{35}}{\frac{56.8}{35}+\frac{80.6}{35}}.100\%=48,3\left(\%\right)\Rightarrow\%m_{Cu}=100-48,3=51,7\left(\%\right)\)