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Chemistry 16 Online
imqwerty (imqwerty):

By the reaction of of carbon & oxygen , a mixture of co & co2 is obtained. What is the composition by mass of the mixture obtained when 20grams of O2 reacts with 12grams of carbon??

OpenStudy (anonymous):

Do you have the chemical formula for this reaction?

imqwerty (imqwerty):

4C + 3O2 -> 2CO2 + 2CO

OpenStudy (anonymous):

So first you need to convert 20g of O2 into moles.

OpenStudy (anonymous):

And then the 12g of C into moles.

imqwerty (imqwerty):

we get 1mole of C reacting with 0.625 moles of O2

imqwerty (imqwerty):

i think that the limitating reagent concept apply here

OpenStudy (anonymous):

Yep that's it.

OpenStudy (anonymous):

Do you know how the calculation works?

OpenStudy (anonymous):

@imqwerty

imqwerty (imqwerty):

1st i got the limitating reagent as O2 2nd - according to the equation 3 moles of O2 give 2moles of CO2 and 2 moles of CO. 3rd - we have 20/32 moles of O2 so we will get 5/12 moles of both co and co2 4th - we have the moles of co and co2 nw we calculate their mass 5th - mass of co = 35/3gm and mass of co2 = 55/3 6th mass of co2:mass of co = 11:7 but the answer given is 21:11

OpenStudy (anonymous):

How did you get that ratio?

imqwerty (imqwerty):

mass of co2/mass of co = 55/3 x 3/35 =11:7

OpenStudy (anonymous):

It seems like what your doing is right, and that answer doesn't seem to be right to me, as they got 21g for CO2 nO2 = m/M = 20g/32g/mol = 0.625mol nCO2 = nO2 x 2/3 = 0.41667 mol mCO2 = M x m = 0.41667 mol x 44.01g/mol = 18.333 g = 55/3 They used the mass of O2 needed rather than the actual mass.

OpenStudy (anonymous):

My ratio 18.338967/11.671767 ~ 1.5714 thats almost the same as 11/7 = 1.5714 So its correct to me, but then again maybe u should double check with your teacher as to why the answer is different.

imqwerty (imqwerty):

thanks @ShizukaTheOtaku

OpenStudy (anonymous):

No problem.

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