What mass of iron(III) chloride is produced when 100.0 mL of 0.100 M solution of barium chloride is reacted with 50.00 mL of 0.150 M solution of iron(III) sulfate?What's the limiting, excess reactant?

Answer 1

To answer this question, let's first write a reaction equation and then balance it: Afterwards, we can see the ratio of barium chloride to FeCl₃ :

Reaction Eqn.:

UNbalanced Eqn: BaCl₂ + Fe₂(SO₄)₃ -> FeCl₃ + BaSO₄

To balance the equation, let's start with the most complicated molecule, .Fe₂(SO₄)₃

and balance the Fe₂(SO₄)₃ with the FeCl₃ :

Fe₂(SO₄)₃ goes to 2 Fe⁺⁺⁺ and 3 SO₄^2-

To form 2 FeCl₃ , we need 6 Cl^-

So we now know PART of the balanced equation as:

Fe₂(SO₄)₃ ....;> 2 FeCl₃

Now the 6 Cl^- must come from 3 BaCl₂ so we have:

3 BaCl₂ + Fe₂(SO₄)₃ ....;> 2 FeCl₃

At this point, we probably already have enough information to answer the question. But for completeness, let's finish the balancing:

3 BaCl₂ + Fe₂(SO₄)₃ ....;> 2 FeCl₃ + 3 BaSO₄

Now we can finally determine the limiting information from the initial volumes and concentrations:

BaCl₂: 100 mL of 0.100 M means (0.100 L)(0.100 mol/L)( 2 mol of Cl^- per mol of BaCl₂) = 0.0200 mol Cl^-

Fe₂(SO₄)₃ : (0.0500 L)(0.150 mol/L)(2 mol Fe³⁺ per mol Fe₂(SO₄)₃ ) = 0.0150 mol Fe³⁺.

To completely react the Fe³⁺, we would need (0.0150)(3) = 0.0450 mol Cl^-. But we only have 0.0200 mol !

Therefore the Cl^- must be the limiting species, and BaCl₂ must be the limiting reagent, and the Fe₂(SO₄)₃ must be the excess reagent.