Question

For the following reaction, 47.8 grams of iron are allowed to react with 83.8 grams of chlorine gas. iron (s) + chlorine (g) iron(III) chloride (s) What is the maximum amount of iron(III) chloride that can be formed? grams What is the FORMULA for the limiting reagent? SubScriptSuperScript Help What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer:

(a) The limiting reagent is the $Cl_{2}$

(b) $127.8gFeCl_{3}$ are formed

(c) $3.8gFe$ remain after the reaction is complete

Explanation:

First we are going to write down the balanced equation:

$_{2}Fe(s)+_{3}Cl_{2}(g)=_{2}FeCl_{3}(s)$

Then we are going to find the number of moles of each reactant:

-For Fe:

$47.8gFe*\frac{1molFe}{55.845gFe}=0.86molesFe$

-For $Cl_{2}$:

$83.8gCl_{2}*\frac{1molCl_{2}}{70.90gCl_{2}}=1.18molesCl_{2}$

Then we are going to divide the number of moles by the stoichiometric coefficient for each reactant to find the limiting reagent:

-For Fe:

$\frac{0.86}{2}=0.43$

-For $Cl_{2}$:

$\frac{1.18}{3}=0.39$

So, the limiting reagent is the $Cl_{2}$.

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:

$83.8gCl_{2}*\frac{1molCl_{2}}{70.90gCl_{2}}*\frac{2molesFeCl_{3} }{3molesCl_{2}}*\frac{162.2gFeCl_{3}}{1molFeCl_{3}}=127.8gFeCl_{3}$

Finally we are going to calculate the amount of the excess reagent that remains after the reaction is complete:

$127.8gFeCl_{3}*\frac{1molFeCl_{3}}{162.2gFeCl_{3}}*\frac{2molesFe}{2molesFeCl_{3}}*\frac{55.84gFe}{1molFe}=44gFe$

$47.8gFe-44.0gFe=3.8gFe$