PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\)
a 2a a
\(Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\)
2a 4a 2a 2a
\(n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
\(\Rightarrow2a+4a=0,8\Leftrightarrow6a=0,8\Leftrightarrow a=\dfrac{2}{15}\)
\(\Rightarrow n_{Fe}=\dfrac{2}{15}.2=\dfrac{4}{15}\left(mol\right)\)
\(m_{hh}=m_{Fe}+m_{Zn}=\dfrac{4}{15}.56+\dfrac{2}{15}.65=23,6>17,7\)
\(\Rightarrow\)Kim loại phản ứng hết, axit dư
Ta có a là nZn, 2a là nFe nên:
\(65a+56.2a=17,7\)
\(177a=17,7\Rightarrow a=0,1\left(mol\right)\)
\(\Rightarrow n_{Fe}=0,1.2=0,2\left(mol\right)\)
\(\sum n_{H_2}=a+2a=0,1+0,2=0,3\left(mol\right)\)
\(\sum V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(\%Fe=\dfrac{m_{Fe}}{m_{hh}}=\dfrac{0,2.56}{17,7}\simeq63,28\%\)
\(\%Zn=100\%-63,28\%=36,72\%\)
