Limiting Reactants: Compare Product Method (Mass/Mass)

Question

For the following reaction, 21.5 grams of iron are allowed to react with 33.7 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?

What amount of the excess reagent remains after the reaction is complete?  grams

Answer 1

The method of comparing product requires you to calculate the mass of product that is produced from each reactant, and seeing which reactant gives the least amount of product. That will be the limiting reactant.

2Fe(s) + 3Cl₂(g) ==> 2FeCl₃(s) ... balanced equation

For Fe: 21.5 g Fe x 1 mol Fe/55.8 g = 0.385 mol Fe x 2 mol FeCl₃/2 mol Fe x 162 g FeCl₃/mol = 62.4 g FeCl₃

For Cl₂: 33.7g Cl₂ x 1 mol Cl₂/70.9g= 0.475 mol Cl₂ x 2 mol FeCl₃/3 mol Cl₂ x 162g FeCl₃/mol = 51.3 g FeCl₃

Thus, Cl₂ is limiting

The maximum amount of FeCl3 that can be formed = 51.3 g FeCl₃

To find amount of excess reagent (Fe) remaining, we find how much was used up, and then subtract that from how much we started with.

moles Fe used up = 0.475 mol Cl2 x 2 mol Fe / 3 mole Cl2 = 0.317 moles Fe used

moles Fe remaining = 0.385 mol - 0.317 mol = 0.068 moles remaining

grams Fe remaining = 0.068 moles x 55.8 g/mol = 3.79 g Fe remain