a) Fe + 2HCl → FeCl2 + H2 (1)
Zn + 2HCl → ZnCl2 + H2 (2)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Gọi \(x,y\) lần lượt là số mol của Fe, Zn
Ta có: \(\left\{{}\begin{matrix}56x+65y=9,3\\x+y=0,15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,05\times56=2,8\left(g\right)\)
\(m_{Zn}=0,1\times65=6,5\left(g\right)\)
c) Theo PT1,2: \(\Sigma n_{HCl}=2\Sigma n_{H_2}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{10,95}{10\%}=109,5\left(g\right)\)
d) \(\Sigma m_{dd}saupư=9,3+109,5=118,8\left(g\right)\)
Theo PT1: \(n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,05\times127=6,35\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{6,35}{118,8}\times100\%=5,35\%\)
Theo PT2: \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,1\times136=13,6\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{118,8}\times100\%=11,45\%\)
