For the reaction ? C6H6 + ? O2 → ? CO2+? H2O , what is the maximum amount of CO2 which could be formed from 11.43 g of C6H6 and 5.93 g of O2? Answer in units of mol.

Answer 1

Hello Tapasiya,

I'm not certain which aspect of this problem you'd like help on. I'll summarize the steps, but feel free to ask if you'd like more detail on a portion of the process. Balancing the equation was a little tricky, in part die to the fact that the source of oxygen (O₂) comes in pairs, so even amounts, which the products take up only 1 at a time (by the H₂O). So I had to make a couple of iterations to arrive at:

2C6H6 + 15O2 = 12CO2 + 6H2O

Calculate the moles of the reactants by dividing the grams by their molar masses (g/mole).

C6H6: 11.5g/78g/mole = 0.1449 moles

O2: 5.93 grams/32g/mole = 0.185 moles

Uh oh . . . Although we have slightly more moles of O2 than C6H6, the balanced equation says we need (15/2), or 7.5 times more moles than C6H6 (1.087 moles is needed, compared to the 0.185 moles provided.. Not happening, so O2 must be a limiting reagent. Once it is depleted, the reaction stops, leaving excess C6H6 behind. We'll assume we use all the oxygen to produce CO2, so we'll base our calculations on that.

Assume the 0.1853 moles of O2 is completely consumed. That will require (2 moles C6H6/15 moles O2) *(0.1852 moles O2) = 0.0247 moles of C6H6, leaving 0.1202 moles of the C6H6 unreacted.

Now let's find the amount of CO2 produced from 0.0247 moles of C6H6. The balanced equation says we'll get 12 moles CO2 for every 2 moles C6H6. So 6*(0.0247) = 0.1482 moles CO2.

[You can convert this into grams by multiplying by CO2's molar mass. I get 6.253 grams, but the question wants moles CO2]

I hope this helps,

Bob