Bài 13:
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
⇒ 27x + 56y = 19,3 (1)
Ta có: \(n_{H_2}=0,65\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=0,65\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,3.27=8,1\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
c, Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)
⇒ mAlCl3 + mFeCl2 = 0,3.133,5 + 0,2.127 = 65,45 (g)
