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What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate

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Answer:

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

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