Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: rambo007 on January 28, 2025, 04:51:05 AM
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Hi, is this question incomplete or I don't understand the question？
When (9.960x10^-1) g of silver nitrate is dissolved in water and then mixed with excess potassium iodide solution (6.00x10^-1) g of precipitate form. What is the percent yield of the reaction? (Answer to 3 s.d. in proper scientific notation.)
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You need solubility product of the precipitate.
Convert both mass to mole
Calculate what is in excess.
The remaining divided to the input times 100% is the yield.
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Start by writing a balanced equation and finding relevant molecular weights. Show your work, and we can take it from there.
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Hi, is this question incomplete or I don't understand the question？
When (9.960x10^-1) g of silver nitrate is dissolved in water and then mixed with excess potassium iodide solution (6.00x10^-1) g of precipitate form. What is the percent yield of the reaction? (Answer to 3 s.d. in proper scientific notation.)
It looks like the question is missing some details, like the reaction equation and the theoretical yield. To calculate the percent yield, you need to compare the actual amount of precipitate formed with the amount you would have expected if the reaction went perfectly (theoretical yield). If you have the molar masses and the equation, you can easily work it out.
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No, the theoretical yield is given by the solubility product. Of course this has to be looked up.