Đổi:400ml=0,4l
Gọi x;2y là số mol Fe,Al
Theo gt:\(m_{hhKL}\)=\(m_{Fe}+m_{Al}\)=56x+27y.2
=56x+54y=11(1)
Ta có PTHH:
Fe+\(H_2SO_4\)->\(FeSO_4\)+\(H_2\)(1)
x..........x................x.................(mol)
4Al+6\(H_2SO_4\)->2\(Al_2(SO_4)_3\)+3\(H_2\)(2)
2y...........3y...............y.........................(mol)
Ta có:\(C_{MddH_2SO_4}\)=1M
=>\(n_{H_2SO_4}\)=1.0,4=0,4mol
Theo PTHH(1);(2):
\(n_{H_2SO_4}\)=x+3y=0,4(2)
Từ (1);(2)=>\(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m_{Al}=54y=54.0,1=5,4\left(g\right)\\m_{Fe}=56x=56.0,1=5,6\left(g\right)\end{matrix}\right.\)
Theo PTHH(1);(2):\(n_{FeSO_4}\)=x=0,1(mol)
\(n_{Al_2\left(SO_4\right)_3}\)=y=0,1(mol)
Vậy \(C_{M\left(FeSO_4\right)}\)=0,1:0,4=0,25M
\(C_{MAl_2\left(SO_4\right)_3}\)=0,1:0,4=0,25M
\(n_{H_2SO_4}=0,4.1=0,4\left(mol\right)\)
Gọi x, y lần lượt là số mol của Al, Fe
Pt: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)
x \(\rightarrow\dfrac{3x}{2}\) \(\rightarrow0,1mol\)
Pt: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (2)
y \(\rightarrow y\) \(\rightarrow0,1mol\)
(1)(2) \(\Rightarrow\left\{{}\begin{matrix}1,5x+y=0,4\\27x+56y=11\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,4}=0,25M\)
\(C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\)
