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A limiting reactants problem.

6.32g of sodium sulfate is reacted with 12.03g of barium nitrate. How many grams of the precipitate would you expect to collect.

The answer my teacher gave me is 16.0g BaSO4, but I cannot get it.

The equation I formed is:
Na2SO4 + Ba(NO3)2 --> 2NaNO3 + BaSO4

Thanks in advance!

  • Chemistry -

    I think 16.0 g BaSO4 is wrong.
    I work these the long way.
    First we take mols Na2SO4 and all the Ba(NO3)2 needed and solve for mols BaSO4 formed.
    Next we take mols Ba(NO)3 and all the Na2SO4 needed and solve for mols BaSO4 formed.
    The smaller number of mols BaSO4 formed wins.
    mols Na2SO4 = 6.32/142 = 0.0445
    That gives you 0.0445 mols BaSO4 and convert to g BaSO4 = 0.0445*233.39 = 10.4g

    mols Ba(NO3)2 = 12.03/261.337 = 0.046
    0.046*233.39 = 10.7 g
    10.4g BaSO4 is what you will obtain. How that compare with your value?

  • Chemistry -

    I got 10.4g as well. My teacher must have put down the incorrect answer... thank you for your help!

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