a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b, Gọi: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 56x + 27y = 5,5 (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,05.56=2,8\left(g\right)\\m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{2,8}{5,5}.100\%\approx50,91\%\\\%m_{Al}\approx49,09\%\end{matrix}\right.\)
c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,2}{1,2}=\dfrac{1}{6}\left(l\right)\)
d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,05\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\end{matrix}\right.\)
⇒ m muối = mFeSO4 + mAl2(SO4)3 = 0,05.152 + 0,05.342 = 24,7 (g)
Đổi \(4,48dm^3=4,48\left(l\right)\)
\(n_{hh}khí=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Đặt số mol của Fe,Al lần lượt là a,b (mol)
Ta có PTHH:
\(Fe+H_2SO_4->FeSO_4+H_2\)
a --> a a a (mol)
\(2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
b --> 3b/2 b/2 3b/2 (mol)
Theo đề bài ta có hai phương trình
\(\left\{{}\begin{matrix}56a+27b=5,5\\a+\dfrac{3b}{2}=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)
--> \(m_{Fe}=0,05\cdot56=2,8\left(g\right)\)
--> \(m_{Al}=0,1\cdot27=2,7\left(g\right)\)
\(\%m_{Fe}=\dfrac{2,8}{5,5}\cdot100\%\approx50,9\%\)
\(\%m_{Al}=\dfrac{2,7}{5,5}\cdot100\%\approx49,1\%\)
Tổng số mol \(H_2SO_4\) phản ứng là \(a+\dfrac{3b}{2}=0,05+\dfrac{0,1\cdot3}{2}=0,2\left(mol\right)\)
\( V_{H_2SO_4}=\dfrac{0,2}{1,2}=\dfrac{1}{6}\left(l\right)=166,67ml\)
\(n_{FeSO_4}=0,05->m_{FeSO_4}=0,05\cdot152=7,6\left(g\right)\)
\(n_{Al_2\left(SO_4\right)_3}=0,05->m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
Tổng khối lượng muối trong A là \(7,6+17,1=24,7\left(g\right)\)
