A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by addi

Question

A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 4.912 x 10-1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004 x 10-1 M sodium Iodate, NaIO3 Assume that, in the experiment, 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?

Explanation / Answer

Ba(NO3)2 + 2 NaIO3 Ba(IO3)2 + 2 NaNO3

(30.00 mL) x (4.912 x 10^-1 M Ba(NO3)2) = 14.736 mmol Ba(NO3)2
(50.00 mL) x (9.004 x 10^-1 M NaIO3) = 45.02 mmol NaIO3

1 millimol of Ba(NO3)2 reacts completely with 2 millimol of NaIO3

14.736 millimoles of Ba(NO3)2 would react completely with 14.736 x 2 = 29.472 millimoles of NaIO3 but amount of NaIO3 is more than required so Ba(NO3)2 is the Limiting reagent

Theoretical yield of Ba(NO3)2 = (14.736 mmol Ba(NO3)2) x (1 mol Ba(IO3)2 x 1 mol Ba(NO3)2) x
(505.1476 g Ba(IO3)2*H2O/mol) = 7443.85 mg = 7.444 g Ba(IO3)2*H2O

Mass of product expected to isolate = 7.444 - 6.895 g = 0.549 g

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