\(n_{Fe}=0,012\left(mol\right);n_{Cl2}=0,02\left(mol\right)\)
\(PTHH:2Fe+3Cl_2\rightarrow2FeCl_3\)
______0,012__________0,012
Lập tỉ lệ: \(\frac{n_{Fe}}{2}< \frac{n_{Cl2}}{3}\)
Nên Fe pứ hết
\(\Rightarrow m_{muoi}=0,012.\left(56+35,5.3\right)=1,95\left(g\right)\)
\(2Fe+3Cl2-->2FeCl3\)
\(n_{Cl2}=\frac{0,448}{22,4}=0,02\left(mol\right)\)
\(n_{Fe}=\frac{0,672}{22,4}=0,03\left(mol\right)\)
\(n_{Fe}\left(\frac{0,03}{2}\right)>n_{Cl2}\left(\frac{0,02}{3}\right)\)
-->Fe dư
\(n_{FeCl3}=\frac{2}{3}n_{Cl2}=\frac{2}{15}\left(mol\right)\)
\(m_{FeCl3}=\frac{2}{15}.162,5=21,667\left(g\right)\)
