Question

Magnesium reacts with iron (II) Chloride to form magnesium chloride and iron. A mixture of 41.0g of magnesium and 175 g of iron (II) chloride is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant.

Answer 1

Molar mass of Mg = 24.31 g/mol

mass(Mg)= 41.0 g

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(41.0 g)/(24.31 g/mol)

= 1.687 mol

Molar mass of FeCl2,

MM = 1*MM(Fe) + 2*MM(Cl)

= 1*55.85 + 2*35.45

= 126.75 g/mol

mass(FeCl2)= 175.0 g

number of mol of FeCl2,

n = mass of FeCl2/molar mass of FeCl2

=(175.0 g)/(126.75 g/mol)

= 1.381 mol

Balanced chemical equation is:

Mg + FeCl2 ---> Fe + MgCl2

1 mol of Mg reacts with 1 mol of FeCl2

for 1.6865 mol of Mg, 1.6865 mol of FeCl2 is required

But we have 1.3807 mol of FeCl2

so, FeCl2 is limiting reagent

we will use FeCl2 in further calculation

According to balanced equation

mol of Mg reacted = (1/1)* moles of FeCl2

= (1/1)*1.3807

= 1.381 mol

mol of Mg remaining = mol initially present - mol reacted

mol of Mg remaining = 1.6865 - 1.3807

mol of Mg remaining = 0.3059 mol

Molar mass of Mg = 24.31 g/mol

mass of Mg,

m = number of mol * molar mass

= 0.3059 mol * 24.31 g/mol

= 7.44 g

FeCl2 is limiting reactant

mass of excess reactant = 7.44 g