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?

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

FREE Expert Solution
FREE Expert 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)


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Problem Details

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?

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

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.