How many grams of FeCl2 can be produced when 4.5 g of Fe reacts with 5.5 g of Cl2?

Question

How many grams of FeCl2 can be produced when 4.5 g of Fe reacts with 5.5 g of Cl2?

Answer 1

The maximum amount produced will be from the reaction of all of the limiting reagent.

To work out which is the limiting reagent first work out the number of moles of each reactant started with use the equation.

moles = mass / molecular weight

molecular weight Fe = 55.85 g/mol

moles Fe = 4.5 g / 55.85 g

= 0.0806 moles of Fe

molecular weight Cl2 = (2 x 35.45) = 70.9 g/mol

moles Cl2 = 5.5 g / 70.9 g/mol

= 0.0776 moles of Cl2

The balanced equation for the reaction is

Fe + Cl2 ------------> FeCl2

This tells us that Fe and Cl2 react in a 1:1 ratio.

Therefore the limiting reagent is Cl2 (0.0776 mol) and the Fe is in excess.

Now, 1 mole of Cl2 reacts to give 1 mole of FeCl2. Therefore if you only have 0.0776 mole of Cl2 then the maximum FeCl2 you can produce is 0.0776 mol

Mass of FeCl2 possible = mole x molecular weight

molecular weight = 55.85 + (2 x 35.45) = 126.75 g/mol

Therfore mass = 0.0776 mol x 126.75 g/mol

= 9.84 g

Therfore 9.84 g of FeCl2 is possible