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2Fe+3Cl2--->2FeCl2
n Fe=22,4/56=0,4(mol)
n Cl2=8,96/22,4=0,4(mol)
0,4/2>0,4/3
--->Fe dư.Tính theo chất hết
Theo pthh
n FeCl3=2/3n Cl2=0,26667(mol)
m FeCl3=0,26667.162,5=43,3(g)
\(n_{Cl_2}=0,4\left(mol\right);n_{Fe}=0,4\left(mol\right)\\\Rightarrow \frac{n_{Cl_2}}{n_{Fe}}=1\Rightarrow\text{ Chỉ tạo muối }Fe\left(II\right)\\ Fe+Cl_2\rightarrow FeCl_2\)
\(\Rightarrow n_{FeCl_2}=0,4\left(mol\right)\Rightarrow m_{FeCl_2}=50,8\left(g\right)\)
