a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)
c, Ta có: 65nZn + 24nMg = 10,1 (1)
Theo PT: \(n_{H_2}=n_{Zn}+n_{Mg}=0,25\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Mg}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\)
\(A.Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ B.n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Zn}=a,n_{Mg}=b\\ \Rightarrow\left\{{}\begin{matrix}65a+24b=10,1\\a+b=0,25\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,15\\ n_{HCl}=0,1.2+0,15.2=0,5mol\\ C_{M_{HCl}}=\dfrac{0,5}{0,1}=5M\\ C.m_{Zn}=0,1.65=6,5g\\ m_{Mg}=10,1-6,5=3,6g\)
