Chemistry Practice Problems Solution Stoichiometry Practice Problems Solution: What mass of barium sulfate can be produced when 1...

🤓 Based on our data, we think this question is relevant for Professor Littlejohn's class at SAC.

Solution: What mass of barium sulfate can be produced when 100.0 mL of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?

Problem

What mass of barium sulfate can be produced when 100.0 mL of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?

Solution

We’re being asked to calculate the mass of barium sulfate (BaSO4) formed in the reaction of 100.0 mL of 0.100 M BaCl2 solution and 100.0 mL of 0.100 M Fe2(SO4)3.


For this problem, we need to do the following steps:

Step 1: Determine the products of the reaction.

Step 2: Write and balance a chemical equation for the given reaction.

Step 3: Determine the limiting reactant and calculate the mass of BaSO4 produced.



Step 1: Since BaCl2 and Fe2(SO4)3 are ionic compounds, they form ions when dissociating in water. The dissociation of the two compounds is as follows:


For BaCl2: 

The chloride ion, Cl, has a charge of –1. Barium is from Group 2 so it has a charge of +2:

BaCl2(aq)  Ba2+(aq) + 2 Cl(aq)


For Fe2(SO4)3: 

The sulfate ion, SO42–, has a charge of –2. Iron then has a charge of +3:

Fe2(SO4)3(aq)  2 Fe3+(aq) + 3 SO42–(aq)


Solution BlurView Complete Written Solution