A student synthesized 6.895 g of barium iodate monohydrate. Ba(lO_3)2*H_2 O by a

Question

A student synthesized 6.895 g of barium iodate monohydrate. Ba(lO_3)2*H_2 O by adding 30.00 mL of 5.912 times 10^-1 M barium nitrate. Ba(NO_3)_2- to 50.00 mL of 9.004 times 10^-1 M sodium Iodate. NalO_3 Calculate the percent yield of barium iodate monohydrate the student obtained in this experiment. When reviewing the procedure the student found that 4.912 times 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(lO_3)_2'H_20 by adding 30.00 mL of 5.912 times 10^-1 M barium nitrate, Ba(NO_3)_2, to 50.00 mL of 9.004 times 10^-1 M sodium Iodate. NalO_3 Calculate the theoretical yield of barium iodate monohydrate using 30.00 mL of 4.912 times 10^-1 M barium nitrate solution 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? Note* leave your answer unit-less. I will assume your answer is in grams. For example if your answer is 2 g enter 2 in the answer box

Explanation / Answer

(1)

Reaction taking place is:

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

After this 1 molecule of water gets attached to 1 molecule of Ba(IO3)2

Moles of barium nitrate taken = Molarity*Volume = 0.5912*0.3 = 0.17736

Moles of sodium iodate taken = Molarity*Volume = 0.9004*0.5 = 0.4502

Here, barium nitrate is the limiting reagent.

So, moles of barium iodate produced theoretically = 0.17736

So, moles of barium iodate monohydrate produced theoretically = 0.17736

So, theoretical mass produced = 0.17736*505.14 = 89.59 g

Mass actually obtained = 6.895

So, % yield = (6.895/89.59)*100 = 7.696%

(2)

In this case,

Moles of barium nitrate taken = Molarity*Volume = 0.4912*0.3 = 0.14736

So, moles of barium iodate produced theoretically = 0.14736

So, moles of barium iodate monohydrate produced theoretically = 0.14736

So, theoretical mass produced = 0.14736*505.14 = 74.43 g

Mass actually obtained = 6.895

So, % yield = (6.895/74.43)*100 = 9.26%

(3)

Kindly provide the theory for this part because it is not clear what reaction is taking place.

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