mFe+Al=mhh-mCu=15,65-3,2=12,45(g)
nH2=8,4/22,4=0,375(mol)
2Al+3H2SO4--->Al2(SO4)3+3H2
x_____3/2x________x_______3/2x
Fe+H2SO4--->FeSO4+H2
y____y________y_____y
Hệ pt:
\(\left\{{}\begin{matrix}27x+56y=12,45\\1,5x+y=0,375\end{matrix}\right.\Rightarrow x,y\)
=>m Al=>%mAl
=>%mFe
=>%mCu
c)=>\(\Sigma nH2SO4\)
=.mH2SO4
\(\text{a) }pthh:\overset{2Al}{x}\overset{\overset{+3H_2SO_4\overset{t^0}{\rightarrow}Al_2\left(SO_4\right)_3+}{ }}{\overset{ }{ }}\overset{3H_2}{1,5x}\left(1\right)\\ \overset{Fe}{y}\overset{+H_2SO_4\overset{t^0}{\rightarrow}FeSO_4+}{\overset{ }{\overset{ }{ }}}\overset{H_2}{y}\left(2\right)\\ Cu+H_2SO_4\rightarrow không\text{ }phản\text{ }ứng\)
b) Khi cho \(Al;Cu;Fe\) tác dụng \(H_2SO_4\) thì \(Cu\) không phản ứng
\(\Rightarrow m_{Cu}=3,2\left(g\right)\\ \Rightarrow m_{Al}+m_{Fe}=15,65-3,2=12,45\left(g\right)\)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)
Từ \(\left(1\right)\) và \(\left(2\right),\) ta có hệ phương trình \(:\left\{{}\begin{matrix}1,5x+y=0,375\\27x+56y=12,45\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,15\end{matrix}\right.\)
\(\Rightarrow n_{Al}=x=0,15\left(mol\right)\Rightarrow m_{Al}=n\cdot M=0,15\cdot27=4,05\left(g\right)\\ n_{Fe}=y=0,15\left(mol\right)\Rightarrow m_{Fe}=n\cdot M=0,15\cdot56=8,4\left(g\right)\\ \Rightarrow\%Cu=\dfrac{3,2\cdot100}{15,65}=20,45\%\\ \%Fe=\dfrac{8,4\cdot100}{15,65}=53,67\%\\ \%Cu=\dfrac{4,05\cdot100}{15,65}=25,88\%\)
c) Theo \(pthh\left(1\right):n_{H_2SO_4\left(1\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,15=0,225\left(mol\right)\)
Theo \(pthh\left(2\right):n_{H_2SO_4\left(2\right)}=n_{Fe}=0,15\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}=n_{H_2SO_4\left(1\right)}+n_{H_2SO_4\left(2\right)}=0,225+0,15=0,375\left(mol\right)\\ \Rightarrow\Rightarrow\Sigma m_{H_2SO_4}=\Sigma n_{H_2SO_4}\cdot M=0,375\cdot98=36,75\left(g\right)\)
