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.
To determine the mass of product expected to be isolated, we need to consider the solubility of barium iodate monohydrate in both 25°C water and 4°C water.
Given:
- Solubility of barium iodate monohydrate in 25°C water is 0.028 g per 100 mL of water.
- Solubility of barium iodate monohydrate in 4°C water is 0.010 g per 100 mL of water.
First, we need to find out how much barium iodate monohydrate can dissolve in the 30.00 mL of 25°C water.
Step 1: Convert the volume of water to mL:
30.00 mL
Step 2: Calculate the amount of barium iodate monohydrate that can dissolve in 30.00 mL of 25°C water.
(0.028 g / 100 mL) * 30.00 mL = 0.0084 g
Therefore, 0.0084 g of barium iodate monohydrate can dissolve in 30.00 mL of 25°C water.
Now, we'll compare this calculated value (0.0084 g) with the theoretical yield of barium iodate monohydrate obtained by using 30.00 mL of 4.912 x 10^-1 M barium nitrate solution.
Unfortunately, the information provided doesn't specify the relationship between the concentration of the barium nitrate solution and the yield of barium iodate monohydrate. Without this information, we are unable to determine the mass of product expected to be isolated using 30.00 mL of 4.912 x 10^-1 M barium nitrate solution.
If you have any additional information or clarification about the relationship between the concentration of barium nitrate solution and product yield, please provide it for a more accurate answer.