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sara1234
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Does anyone have Chemistry in Connections i need help with a percent yeild lab
 one year ago
 one year ago
sara1234 Group Title
Does anyone have Chemistry in Connections i need help with a percent yeild lab
 one year ago
 one year ago

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Australopithecus Group TitleBest ResponseYou've already chosen the best response.3
\[Precent Yield = (\frac{Moles of Acquired Product}{MolesOfTheoreticalYield})100\]
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.3
I don't mind helping you with your lab if you show me the reaction you conducted, the yield of product you had and the amount of reagents you used
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.3
Just use the formula \[Moles = \frac{Grams}{Molecular Mass}\] to convert to moles
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.3
You can just give me your reaction mechanism and I can show you the general idea on how to calculate the theoretical yield. All you have to do is figure out the moles of desired product produced from the moles of the limiting reagent you added in the reaction. Then just convert the grams of product you got from the reaction to moles and use the formula above
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.3
I can just give you an example I guess Just say you performed this reaction: H2SO4 + 2NaOH > 2H2O + Na2SO4 and you got a yield of 5.0 grams of sodium sulphate (Na2SO4) and you reacted 10mL 18M H2SO4 and 20grams of NaOH so first convert both to moles To convert H2SO4 to moles we use the formula Molarity = Moles/Liters so 18M = x moles/0.01L 18M*0.01 = 0.18mol Now convert NaOH to moles, using the formula I posted above 20g/(22.989g/mol + 15.9994g/mol + 1.00794g/mol) = 0.5mol NaOH so we have, 0.5moles of NaOH and 0.18moles of H2SO4 ok now we figure out what the limiting reagent is by converting both to moles of the desired product, what ever gives us the least amount of moles product is our limiting reagent Lets find out if NaOH is our limiting reagent look at reaction again H2SO4 + 2NaOH > 2H2O + Na2SO4 we are going from NaOH to Na2SO4, note the coefficients, we multiply by 2 and divide by 1 ( (0.5moles)(2) )/1 = 1mol of Na2SO4 Now lets find out if H2SO4 is our limiting reagent H2SO4 + 2NaOH > 2H2O + Na2SO4 Same deal, this time though we multiply by 1 and divide by 1, (this is a 1 to 1 reaction) ( (0.18moles)(1) )/1 = 0.18mol of Na2SO4 we see that the H2SO4 gives us the smallest amount of product thus it is the limiting reagent when it is used up the reaction ends (In this case it is an equilibrium reaction though but dont worry about that) Thus we have a theoretical yield of 0.18mol of Product Now we just need to find moles of product obtained 5.0 grams/(142.02g/mol) = 0.0352mol so we got 0.0352mol of product and a theoretical yield of 0.18mol now we just apply the formula I stated in my first post to find the percent yield (0.0352mol/0.18mol)*100 = 19.6%
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.3
I made a mistake here look at reaction again H2SO4 + 2NaOH > 2H2O + Na2SO4 we are going from NaOH to Na2SO4, note the coefficients, we divide by 2 and multiply by 1 This is because for every 2 NaOH we get 1 Na2SO4 ( (0.5moles)(1) )/2 = 0.25mol of Na2SO4
 one year ago
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