Chemistry

When reviewing the procedure the student found that 4.912 x 10-1 M barium nitrate solution had been used instead of that in the following procedure:

A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 5.912 x 10-1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004 x 10-1 M sodium Iodate, NaIO3

Calculate the theoretical yield of barium iodate monohydrate using 30.00 mL of 4.912 x 10-1 M barium nitrate solution

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asked by Steve
  1. Ba(NO3)2 + 2NaIO3 + H2O ==> Ba(IO3)2*H2O + 2NaNO3
    Ba(NO3)2 is the limiting reagent with either number.
    mols Ba(NO3)2 = M x L = 0.4912 x 0.030 = 0.014736 = mols Ba(IO3)2.H2O
    g Ba(IO3)2 = mols x molar mass = theoretical yield (TY). Actual yield (AY) = 6.895g in the problem.
    % yield (but not asked for above) = (AY/TY)*100 = ?

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