Following the general procedure described in this experiment, a student synthesized 6.895 g of barium iodate monohydrate,
Question:
Following the general procedure described in this experiment, a student synthesized 6.895 g of barium iodate monohydrate, Ba(IO 3 ) 2. H 2 ), by adding 30.00 mL of 5.912x10 -1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004x10 -1 M sodium iodate, NaIO 3 .
(A) Write the chemical equation for the reaction of solutions of barium nitrate and sodium iodate.
(B) Calculate the percent yield of barium iodate monohydrate the student obtained in this experiment. When reviewing the procedure and calculations, the student discovered that a 4.912x10 -1 M barium nitrate solution had been used instead of a 5.912x10 -1 M solution.
(C) Calculate the percent yield of barium iodate monohydrate using 30.00 mL of 4.912x10 -1 M barium nitrate solution and 50.00 mL of 9.004x10 -1 M sodium iodate solution.
(D) Calculate the present error in the present yield calculated in (2) compared with the correct percent yield calculated in (C) 2. Assume that, in the experment described in question (1), 125 mL of 25 C distilled water was used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (at4 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.(1) What mass of product would you expect to isolate? (2) Calculate the percent error as a result of using 125 mL of 25 C water, compared with the correct yield using 20 mL of 4 C water
Probability and Statistics
ISBN: 978-0321500465
4th edition
Authors: Morris H. DeGroot, Mark J. Schervish