\(PTHH:\text{Cu+4HNO3}\rightarrow\text{Cu(NO3)2+2NO2+2H2O}\)
\(\text{Fe+2HCl}\rightarrow\text{FeCl2+H2}\)
\(\text{2Al+6HCl}\rightarrow\text{2AlCl3+3H2}\)
Gọi a, b lần lượt là số mol của Fe và Al
Ta có :
\(\left\{{}\begin{matrix}\text{nNO2=0,1(mol)}\\\text{nH2=0,175(mol)}\end{matrix}\right.\)
=>nCu=0,05(mol)
=>mCu=0,05x64=3,2(g)
\(\Rightarrow\left\{{}\begin{matrix}\text{56a+27b=6,95}\\\text{a+3b/2=0,175}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\text{a=0,1}\\\text{b=0,05}\end{matrix}\right.\)
\(\text{%Cu=3,2/10,15x100=31,53%}\)
\(\text{%Fe=0,1x56/10,15x100=55,17%}\)
\(\text{%Al=13,3%}\)
Cu + 4HNO3 -----> Cu(NO3)2 + 2NO2 + 2H2O
Fe+2HCl---->FeCl2+H2
2Al+6HCl---->2AlCl3+3H2
n\(_{NO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo pthh1
n\(_{Cu}=\frac{1}{2}n_{NO2}=0,05\left(mol\right)\)
%m Cu=\(\frac{0,05.64}{10,15}.100\%=31,53\%\)
m\(_{Fe}+m_{Al}=10,15-3,2=6,95\left(g\right)\)
n\(_{H2}=\frac{3,92}{22,4}=0,175\left(mol\right)\)
Gọi n\(_{Fe}=x,n_{Al}=y\)
Theo bài ra ta có pt
\(\)\(\left\{{}\begin{matrix}56x+27y=6,95\\x+1,5x=0,175\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
%m\(_{Fe}=\frac{0,1.56}{10,15}.100\%=55,17\text{%}\)
% m Al=100-55,17-31,53=13,3%
