What mass of barium sulphate 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) sulphate?

Please explain!

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  1. moles BaCl2 = M x L = 0.100 x 0.100L = ?
    moles Fe2(SO4)3 = M x L = ?
    ..3BaCl2 + Fe2(SO4)3 ==> 3BaSO4 + 2FeCl3

    You have the initial amount from above.
    Convert mole BaCl2 to moles BaSO4
    Convert moles Fe2(SO4)3 to mols BaSO4.
    The SMALLER number is ALWAYS the number to choose when you have a limiting reagent problem (and we know it is a limiting reagent because amounts for BOTH reactants are given).
    Then grams = moles BaSO4 x molar mass BaSO4.

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  2. Okay this makes sense thank you!

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