# If I do this reaction with 35 grams of C6H10 and 45 grams of oxygen, how many grams of carbon dioxide will be formed?

## 1 Answers

John McCann answered
First the equation!! ( you are given two masses to find a limiting reactant )

2C6H10 + 17O2 -> 12CO2 + 10H2O

Find limiting reactant; get mole of reactants first

35 grams C6H10 (1 mole C6H10/82.14 grams) = 0.4261 moles C6H10

45 grams O2 (1 mole O2/32.0 grams) = 1.406 mole O2

I suspect C6H10 limits; check

0.4261 moles C6H10 (17 mole O2/2 moles C6H10) = 3.622 moles O2

you have a lot more O2 than that; C6H10 limits and will drive the reaction ( check the other way around and you see that C6H10 does not make the cut )

0.4261 moles C6H10 (12 moles CO2/2 moles C6H10)(44.01 grams/1 mole CO2)

= 113 grams CO2 produced
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thanked the writer.
Oddman commented
Maybe I'm missing something, but it seems to me that 3.622 moles of O2 needed to completely use the C6H10 is a lot more than the 1.406 moles of O2 available.
Oddman commented
As a check, you have reactants totaling 80 grams producing products, one of which is reportedly 113 grams. A lot of energy must have come from somewhere to create more than 33 grams of mass.
John McCann commented
You are right. I made a big blunder here. O2 limits

1.406 moles O2 (2 mole C6H10/17 mole O2) = 0.1654 mole C5H10---which is more than enough.--- O2 drives the reaction.

1.406 moles O2 (12 moles CO2/17 moles O2)(44.01 grams/1 mole CO2)

= 44 grams of CO2 produced
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