Posted by Steve on .
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
HERE IS THE QUESTION: SUPPLEMENTAL INFO IS ABOVE
Assume that, in the experiment described in Post-Laboratory Question 2, 125 mL of 25 C distilled water was used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (4 C).
The solubility of Barium iodate monohydrate in 25 C water is 0.028 g per 100 mL of water; in 4 C water, it is 0.010 g per 100 mL of water.
What mass of product would you expect to isolate?
See your other post above.