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Topic: Percent yield of barium sulfate  (Read 861 times)

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Offline PanpanBamboo

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Percent yield of barium sulfate
« on: April 16, 2020, 03:54:32 PM »
I have a question on my homework asking for the percent yield in a solution of barium II chloride and potassium sulfate. The concentration of each of the reactants is 0.15 M for barium II chloride and 0.24 M for potassium sulfate. 115.0 mL of each are allowed to react, and 5.1418 grams of solid are created.
I got that the solid created is barium sulfate, and it is in a 1:1 ratio with all the reactants in the net ionic equation. However, no matter how many times I try I always get that the actual yield of 5.1418 grams is higher than the theoretical yield when I find the limiting reactant and multiplying it by the molar mass.
Any ideas? Thanks!

Offline AWK

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Re: Percent yield of barium sulfate
« Reply #1 on: April 16, 2020, 04:13:07 PM »
Nice yield - 127 %.
Some numbers are wrong.
AWK

Offline PanpanBamboo

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Re: Percent yield of barium sulfate
« Reply #2 on: April 16, 2020, 04:18:53 PM »
Thanks for the attempt. I also got a yield of 127%. Here are the quotes from the question:
"reaction of 0.15 M barium (II) chloride, with 0.24 M potassium sulfate"
"Determine the percent yield if 115.0 mL of each reactant were allowed to react, and a mass of 5.1418 g of solid were obtained."

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