For the reaction ?C6H6+ ?O2 -> ?CO2 + ?H2O 47.6 grams of C6H6 are allowed to react with 136.8 grams of O2. How much CO2 will be produced by this reaction? ( answer in grams)

Answer 1

Hello, Claudia,

We need a balanced equation to start. I believe this works:

2C6H6 + 15O2 = 12CO2 + 6H2O

Convert the masses of C6H6 and O2 into moles by dividing their masses by their molar masses.

C6H6 = (47.6g/78g/mole) = 0.610 moles C6H6

O2 = 136.8 g/32 g/mole) = 4.275 moles

Now determine if one of the two reactants is a limiting reagent (i.e., there isn't enough of one to completely consume the other).

The molar ratio of O2 to C6H6 is (15/2) or 7.5. Multiply this by the number of moles C6H6 and we'll have the number of moles of O2 needed to combust all the C6H6. We find that 7.5 x 0.61 moles = 4.58 moles O2 required. We have less than that (4.28 moles O2), so O2 will limit the amount of C6H6 consumed. So we must bae our calculations on the moles O2 to determine the maximum amount of CO2 produced. (Some C6H6 will remain after all the O2 is consumed.

The molar ratio of CO2 to O2 is (12/15) or 0.8. We get 0.8 moles of CO2 per mole O2 consumed. Multiply 0.8 times the moles O2 to calculate the moles CO2 produced before the O2 runs out.

(0.8)*(4.28 moles O2) = 3.42 moles CO2. Convert this into grams by multiplying by CO2's molar mass: (3.42 moles CO2)*(44 g CO2/mole CO2) = 150.5 grams CO2

I hope this helps,

Bob