Asked by Anonymous
Barium sulfate forms as a precipitate in the following reaction:
Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNo3 (aq)
When 35.0g of Ba(nO3)s is reacted with excess Na2SO4, 29.8g of BaSO4 is recovered by the chemist.
a) Calculate the theoretical yield of BaSO4.
b) Calculate the percentage yield of BaSO4.
note: I'm not sure what I do because two of the grams -35.0g and 29.8g - are confusing me. I tried finding the molar mass. Now what?
The answer for
a) 31.3g
b) 95.2%
But I don't understand how to get it. Please explain, thank you.
Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNo3 (aq)
When 35.0g of Ba(nO3)s is reacted with excess Na2SO4, 29.8g of BaSO4 is recovered by the chemist.
a) Calculate the theoretical yield of BaSO4.
b) Calculate the percentage yield of BaSO4.
note: I'm not sure what I do because two of the grams -35.0g and 29.8g - are confusing me. I tried finding the molar mass. Now what?
The answer for
a) 31.3g
b) 95.2%
But I don't understand how to get it. Please explain, thank you.
Answers
Answered by
Sharina.macasocol
This is my problem too
Answered by
Torah
Please I need full working out in order to understand
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