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 25 mL of a 0.175 M solution of iron(III) sulfate?

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The reaction is BaCl2 + Fe2(SO4)3 → BaSO4 + 2FeCl3. Calculate moles of BaCl2 (0.125 L * 0.150 M = 0.01875 mol) and Fe2(SO4)3 (0.025 L * 0.175 M = 0.004375 mol). BaCl2 is the limiting reactant. Moles of BaSO4 produced = 0.01875 mol. Mass = 0.01875 mol * 233 g/mol = 4.37 g.

What Mass of Barium Sulfate (233 G/mol) Is Produced When