A student synthesized 6.895 g of barium iodate monohydrate by adding 30.00 mL of 5.912 x 10-1 M sodium iodate, NAIO3. 125 mL of distilled water as used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (at 4* C). The solubility of barium iodate monohydrate in 25* C water is .028 g per 100 mL of water; in 4* C water, it is .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, compared with the correct yield using 20 mL of 4* C water.

Please help, thanks!

I can't make sense of the problem, partly because how it is stated and partly because of the sentence structure. I don't know if the 6.895 is the theoretical yield OR if it is the actual yield or it is the actual yield if the chilled water is procedure is followed. Also, it isn't clear if the volume used is 125 or 125+35= 160 mL. I suspect actual yield IF the chilled procedure was followed and 125 mL volume so the following answers are based on those assumptions.

Solubility if the student followed directions for 20 mL chilled water is
0.01g/100mL x (20/100) = 0.002 g Ba(IO3)2 lost due to solubility.

Solubility if the student used 125 mL water at 25 C is
0.028g/100mL x 125/100 = 0.035 g lost due to solubility Ba(IO3)2.

So the student would have collected 6.895g if the chilled water procedure was used (with the assumptions made above). Since 0.035 g was lost, the student actually collected 6.895 - 0.035 = 6.860 g.

%error = (0.035/6.895)*100 = ?

To calculate the mass of product that would be expected to isolate, we would need to consider the reaction that takes place and the limiting reagent.

1) The balanced chemical equation for the reaction is:

Ba(IO3)2 + 2NaI → BaI2 + 2NaIO3

From the reaction stoichiometry, we can see that 1 mole of Ba(IO3)2 reacts with 2 moles of NaI to form 1 mole of BaI2.

To find the number of moles of Ba(IO3)2 in 30.00 mL of 5.912 x 10^(-1) M NaIO3 solution, we can use the formula:

moles = concentration (M) x volume (L)

Converting the volume of NaIO3 solution to liters:

30.00 mL = 30.00 / 1000 L = 0.030 L

Calculating the number of moles of Ba(IO3)2:

moles of NaIO3 = concentration of NaIO3 x volume of NaIO3 solution
= (5.912 x 10^(-1) M) x (0.030 L)

Since the stoichiometric ratio is 1:1 between Ba(IO3)2 and BaI2, the number of moles of Ba(IO3)2 is equal to the number of moles of BaI2 formed.

Given that the molar mass of Ba(IO3)2 is 473.86 g/mol, we can calculate the mass of BaI2 formed:

mass of BaI2 = moles of BaI2 × molar mass of BaI2

2) To calculate the correct yield, we need to consider the solubility of barium iodate monohydrate in both 25 °C and 4 °C water.

The solubility of Ba(IO3)2 monohydrate is given as 0.028 g per 100 mL of water at 25 °C and 0.010 g per 100 mL of water at 4 °C.

First, we calculate the maximum mass of Ba(IO3)2 monohydrate that could be dissolved in each volume of water:

For 125 mL of 25 °C water:
maximum mass = (0.028 g/100 mL) × (125 mL) = 0.035 g

For 20 mL of 4 °C water:
maximum mass = (0.010 g/100 mL) × (20 mL) = 0.002 g

The percent error can be calculated using the formula:

percent error = (actual yield - theoretical yield) / theoretical yield × 100%

In this case, the actual yield is the mass of product obtained and the theoretical yield is the maximum mass of product that could be obtained.

I hope this explanation helps you understand how to calculate the mass of product and the percent error in this given scenario.