tackle this chem question!

[atlanta213]

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2MnO4- + 10I- + 16H+ -> 2MN2+ + 5I2 + 8H20

If 150mg of KI is reacted and 108.5mg of I2 is produced, what is the percent yield for the reaction?

For your reference, this is one of kaplan question.

 

[doc3232]

94.62
First converted grams to mols then from I to I2 then found the yield.

 

[dentalknight]

i still cant get the answer, can someone please explain this a little more step by step. i hate these problems!

 

[DRHOYA]

Use the KI mg's as g's of I. So convert 150mg to g......then convert to moles of I. Once you have mols, use stoichoimetric coefficients 10 to 5 to get mols of I. Then convert back to grams. Then take the given (108.5mg) and divide by the theory # (one you just calculated) and multiply by 100. Hope that helps.👍


My problem is doing all the math calcs without a calculator......sometimes #'s aren't even🙁.

 

[doc3232]

sure thing. KI is a salt that completely dissociates into K+ and I- ions. So, you're starting with 0.150g KI/(molar mass of KI, which is 166g KI/mol KI)=9.04e-4 mol of KI, which produces 9.04e-4 mol of K+ and 9.04e-4 mol of I-. Now, we are told that all of the I- reacts, so 9.04e-4 mol reacts. For every 10 moles of I- reacting, 5 moles of I2 should be produced, so there is a 2:1 ratio of moles of I- reacting to moles of I2 being produced. Theoretically, if there was no experimental loss, we would expect half the amount in moles of I2 to be produced as I-. Therefore, we would expect 9.04e-4mol/2=4.52e-4 mol I2 expected. How much do we actually have?
0.1085g I2/(molar mass of I2, which is 254g I2/mol I2)=4.272-4 mol. Obviously, there is some loss due to experimental error.

Percent yield = (actual yield/theoretical yield) x 100%, which would give us--> (4.27/4.52)x100%=94.5% yield

Actually in my O-chem lab, I would always get 100+% yields, I was that good.

 

[atlanta213]

94.5% is answer good job guys!

 

[PooyaH]

Don't we have to convert the moles of each to grams first and then get the percent yield? Isn't is (Mass of Actual / Mass of theoratical) x 100 ?
I know we'd get the same answer, but as long as we know mass should be used instead of the mole! lol

 

[TeamGuo]

Actually in my O-chem lab, I would always get 100+% yields, I was that good.

100% yields... hmmm what kind of experiments were you running... kk

 

[IWantH2O]

Don't we have to convert the moles of each to grams first and then get the percent yield? Isn't is (Mass of Actual / Mass of theoratical) x 100 ?
I know we'd get the same answer, but as long as we know mass should be used instead of the mole! lol

do we always get the same answer...or was it just a coincedence here