a. \(n_{H_2}=\dfrac{6.72}{22,4}=0,3\left(mol\right)\)
PTHH : Mg + 2HCl -> MgCl2 + H2
PTHH : Fe + 2HCl -> FeCl2 + H2
Gọi \(n_{Mg}=a\left(mol\right);n_{Fe}=b\left(mol\right)\)
\(\Rightarrow24a+56b=10,4\left(g\right)\left(1\right)\)
\(\Rightarrow a+b=0,3\left(mol\right)\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrow a=0,2\left(mol\right),b=0,1\left(mol\right)\)
\(\%m_{Mg}=\dfrac{0,2.24}{10,4}=46,1\%\)
\(\%m_{Fe}=100\%-46,1\%=53,9\%\)
b. \(n_{hh}=0,1+0,2=0,3\left(mol\right)\)
Theo PT : \(n_{HCl}=2n_{hh}=0,6\left(mol\right)\)
\(C\%_{HCl}=\dfrac{0,6.36,5}{300}.100=7,3\%\)
c. Theo PT : \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
\(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{10,4+300-0,6}=6,13\%\\C\%_{FeCl_2}=\dfrac{0,1.127}{10,4+300-0,6}=4,09\%\end{matrix}\right.\)