Answer:
(a) The limiting reagent is the [tex]Cl_{2}[/tex]
(b) [tex]127.8gFeCl_{3}[/tex] are formed
(c) [tex]3.8gFe[/tex] remain after the reaction is complete
Explanation:
First we are going to write down the balanced equation:
[tex]_{2}Fe(s)+_{3}Cl_{2}(g)=_{2}FeCl_{3}(s)[/tex]
Then we are going to find the number of moles of each reactant:
-For Fe:
[tex]47.8gFe*\frac{1molFe}{55.845gFe}=0.86molesFe[/tex]
-For [tex]Cl_{2}[/tex]:
[tex]83.8gCl_{2}*\frac{1molCl_{2}}{70.90gCl_{2}}=1.18molesCl_{2}[/tex]
Then we are going to divide the number of moles by the stoichiometric coefficient for each reactant to find the limiting reagent:
-For Fe:
[tex]\frac{0.86}{2}=0.43[/tex]
-For [tex]Cl_{2}[/tex]:
[tex]\frac{1.18}{3}=0.39[/tex]
So, the limiting reagent is the [tex]Cl_{2}[/tex].
Then we are going to find the maximum amount of iron(III) chloride that can be formed, so we take the limiting reagent for the calculations:
[tex]83.8gCl_{2}*\frac{1molCl_{2}}{70.90gCl_{2}}*\frac{2molesFeCl_{3} }{3molesCl_{2}}*\frac{162.2gFeCl_{3}}{1molFeCl_{3}}=127.8gFeCl_{3}[/tex]
Finally we are going to calculate the amount of the excess reagent that remains after the reaction is complete:
[tex]127.8gFeCl_{3}*\frac{1molFeCl_{3}}{162.2gFeCl_{3}}*\frac{2molesFe}{2molesFeCl_{3}}*\frac{55.84gFe}{1molFe}=44gFe[/tex]
[tex]47.8gFe-44.0gFe=3.8gFe[/tex]
