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Chemistry 7 Online
OpenStudy (anonymous):

What is the theoretical yield of Na2SO4, in grams?

OpenStudy (anonymous):

@ryanwhite499 @Hoslos @felipecasal

OpenStudy (anonymous):

what do you think it is..?

OpenStudy (ryanwhite499):

Hold on I'll be back

OpenStudy (anonymous):

Yield, you mean molecular mass?

OpenStudy (anonymous):

@Hoslos : The theoretical yield is the amount predicted by a stoichiometric calculation based on the number of moles of all reactants present.

OpenStudy (anonymous):

Exactly. I asked that because I feel there is lack of information. At least we should hae had the amount of moles reacted in a reaction or something. The statement looks very vague.

OpenStudy (anonymous):

the equation is: H2SO4+2NaOH-> 2H2O+Na2SO4

OpenStudy (ryanwhite499):

This is math @boricua5

OpenStudy (anonymous):

no, its chemistry

OpenStudy (ryanwhite499):

Oh ok. I thought you had to plug the numbers in and stuff

OpenStudy (anonymous):

Yes, you deal with numbers @ryanwhite499 , but this is stoichiometry, part of chemistry responsible for calculations.

OpenStudy (ryanwhite499):

I know @Hoslos. I'm still trying to figure this out. I thought this was a worded math problem at first.

OpenStudy (anonymous):

@boricua5 , I think I need more information, like what amoun reacted, because if you simply give me the reaction, the yield would obviously be 100%, would not be?

OpenStudy (anonymous):

Ok.

OpenStudy (anonymous):

the limiting reactant is 10.8g of Na2SO4

OpenStudy (anonymous):

Ok. Now we are talking. Using your periodic table to find the molecular mass of Na2SO4 is (23*2)+32+(16*4)= 142g. Using ratio of the obtained amount to the expected one is (10.8:142)*100= 7.6%

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