a) \(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b)
\(n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{21,9}{36,5}=0,6\left(mol\right)\)
Lập tỉ số:
\(\frac{0,2}{1}< \frac{0,6}{2}\)
=> PỨ không hết
=> Chọn số mol của Fe
Theo PTHH, ta có:
\(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,2.127=25,4\left(g\right)\)
c)
\(\%H=\frac{M_H.1}{M_{HCl}}.100\%=\frac{1.1}{36,5}.100\%=2,7\%\)
\(\%Cl=100\%-2,7\%=97,3\%\)
a)
nFe=\(\frac{11,2}{56}=0,2\left(mol\right)\);
nHCl=\(\frac{21,9}{36,5}=0,6\left(mol\right)\)
Fe+2HCl\(\rightarrow FeCl_2+H_2\)
1+2 \(\rightarrow1+1\left(mol\right)\)
0,2 + 0,6 \(\rightarrow0,2+0,2\left(mol\right)\)
b)
\(n_{FeCl_2}=0,2\left(mol\right)\)
\(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
c)
%H=\(\frac{1.1}{36,5}\times100\approx2,7\%\)
%Cl=\(\frac{35,5.1}{36,5}\times100\%\approx97,3\%\)
