a student synthesized 6.895g of barium iodate monohydrate, Ba(IO3)2.H2O, by adding 30.00 mL of 5.912*10^-1M barium nitrate, Ba(NO3), to 50.00 mL of 9,004*10^-1 M sodium iodate, NaIO3
(1) write the chemical equation for the reaction of solutions of barium nitrate and sodium iodate
(2) calculate the percent yield of barium iodate monohydrate the student obtained in this experiment
When reviewing the procedure and calculations, the student found that a 4.912*10^-1 M barium nitrate solution had been used instead of a 5.912*10^-1 M solution
(3) calculate the percent yield of barium iodate monohydrate using 30.00 mL of 4.912*10^-1 M barium nitrate solution and 50.00 mL of 9.004*10^-1 M sodium iodate solution.
(4) calculate the percent error in the percent yield calculated in (2), compare with the correct percent yield calculated in (3)
Pleaaaaase, i do really neeeed help
Ba(NO3)2 + 2NaIO3 + H2O ==> Ba(IO3)2.H2O + 2NaNO3
Convert mL and M to moles of each. M x L = moles.
From the moles and the equation, determine which is the limiting reagent and how much of the ppt has been formed. This is the theoretical yield.
% yield = (amount synthesized/theoretical yield)*100 = ??
Follow the same procedure for the second problem.
Post your work if you get stuck.
thank you so much
24
Sure, I'd be happy to help you with your questions. Let's break it down step by step.
(1) To write the chemical equation for the reaction of solutions of barium nitrate and sodium iodate, we need to identify the products formed when these two compounds react.
The chemical formula for barium nitrate is Ba(NO3)2, and the chemical formula for sodium iodate is NaIO3. When these compounds react, they undergo a double displacement reaction.
So, the balanced chemical equation for the reaction is:
Ba(NO3)2 + NaIO3 → Ba(IO3)2 + 2NaNO3
This equation shows that barium nitrate reacts with sodium iodate to form barium iodate and sodium nitrate.
(2) To calculate the percent yield of barium iodate monohydrate, we need to compare the actual yield (6.895g) to the theoretical yield.
The theoretical yield can be calculated using stoichiometry and the balanced chemical equation from step (1). From the equation, we can see that the molar ratio between barium iodate (Ba(IO3)2) and barium nitrate (Ba(NO3)2) is 1:1.
First, let's calculate the number of moles of barium nitrate used:
moles of Ba(NO3)2 = volume (in liters) × molarity
moles of Ba(NO3)2 = 30.00 mL / 1000 mL/L × 5.912 × 10^-1 mol/L
Next, we can calculate the number of moles of barium iodate produced:
moles of Ba(IO3)2 = moles of Ba(NO3)2
Then, we can calculate the theoretical yield of barium iodate in grams:
theoretical yield = moles of Ba(IO3)2 × molar mass of Ba(IO3)2
Finally, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) × 100
(3) Following the same steps as in (2), we can calculate the percent yield of barium iodate monohydrate using 30.00 mL of 4.912 × 10^-1 M barium nitrate solution and 50.00 mL of 9.004 × 10^-1 M sodium iodate solution.
(4) To calculate the percent error in the percent yield calculated in (2) compared to the correct percent yield calculated in (3), we can use the following formula:
percent error = |(experimental value - theoretical value) / theoretical value| × 100
Now, let's plug in the values and calculate the answers.