a student synthesized 6.895g of barium iodate monohydrate, Ba(IO3)2.H2O, by adding 30.00 mL of 5.912*10^-1M barium nitrate, Ba(NO3), to 50.00 mL of 9,004*10^-1 M sodium iodate, NaIO3

Question

(1) write the chemical equation for the reaction of solutions of barium nitrate and sodium iodate
(2) calculate the percent yield of barium iodate monohydrate the student obtained in this experiment

When reviewing the procedure and calculations, the student found that a 4.912*10^-1 M barium nitrate solution had been used instead of a 5.912*10^-1 M solution
(3) calculate the percent yield of barium iodate monohydrate using 30.00 mL of 4.912*10^-1 M barium nitrate solution and 50.00 mL of 9.004*10^-1 M sodium iodate solution.
(4) calculate the percent error in the percent yield calculated in (2), compare with the correct percent yield calculated in (3)

Pleaaaaase, i do really neeeed help

Answer 1

Ba(NO3)2 + 2NaIO3 + H2O ==> Ba(IO3)2.H2O + 2NaNO3

Convert mL and M to moles of each. M x L = moles.

From the moles and the equation, determine which is the limiting reagent and how much of the ppt has been formed. This is the theoretical yield.

% yield = (amount synthesized/theoretical yield)*100 = ??

Follow the same procedure for the second problem.
Post your work if you get stuck.

Answer 2

thank you so much

Answer 3

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