a, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)
⇒ 27x + 24y = 13,2 (1)
Ta có: \(n_{H_2}=0,7\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}=\dfrac{3}{2}x+y=0,7\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,4\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{H_2}=1,4\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{1,4}{2}=0,7\left(l\right)\)
c, Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,4\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{AlCl_3}}=\dfrac{0,4}{0,7}=\dfrac{4}{7}\left(M\right)\\C_{M_{MgCl_2}}=\dfrac{0,1}{0,7}=\dfrac{1}{7}\left(M\right)\end{matrix}\right.\)
Bạn tham khảo nhé!
