Solution:
[tex]P_4+6Cl_2\rightarrow 4PCl_3[/tex]
a) By stoichoimetry If 6 moles of [tex]Cl_2[/tex] gives 4 mole of [tex]PCl_3[/tex] , then 22 moles of [tex]Cl_2[/tex] will give: [tex]\frac{4}{6}\times 22=\frac{44}{3}=14.66 [/tex] moles of [tex]PCl_3[/tex]
b) If 6 moles of [tex]Cl_2[/tex] reacts with one mole of [tex]P_4[/tex] , then 22 moles of [tex]Cl_2[/tex] will react : [tex]\frac{1}{6}\times 22=\frac{11}{3}=3.66[/tex] moles of [tex]P_4[/tex]
Moles of [tex]P_4[/tex] left after the reaction = [tex]5-3.66=\frac{4}{3}=1.34[/tex] moles.
c) 0 Moles of [tex]Cl_2[/tex] , since it is present in less amount it will get completely consumed in the reaction. Hence, it is limiting reagent.
