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