Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced eq....

Question

Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is:

2C4H10(g) + 13O2(g)->8CO2(g) + 10H2O(l)


At 1.00atm and 23*C , how many liters of carbon dioxide are formed by the combustion of 1.00g of butane?

For the above...I have converted 1g of butane=0.017204mol (1g*1mol/58.1234gC4H10) = 0.017204mol C4H10
Then: mol CO2= 1g*1mol/44.011g = 0.2272mol CO2
(if these are the correct initial approach????)

The temp=23*C + 273.15K = 296.15K
The Volume=????
Constant R I will use is: 0.082058L atm mol K
n...?????

I know that I should somehow divide the 8moles of CO2 with the 2moles of C4H10...but not sure how to do this. I tried 0.547L as an answer but it wasn't correct.

Please help and point out what I was doing incorrectly. I thought that I had it figured, but I figured incorrectly.
Thank you in advance!!

Answer 1

Moles C4H10 = 0.0172( you are corret but look at the significant figures)
The ratio between C4H10 and CO2 is 2 : 8
2 : 8 = 0.0172 : x
x = moles CO2 produced = 0.0688
T = 296.16
pV = nRT
V = nRT / V = 0.0688 x 0.0821 x 296.16 / 1.00 = 1.67 L

Answer 2

Gas Cigarette Lighters

Answer 3

Let butane be called B.

1.00gB x 1molB/58gB x 8molCO2/2molB x 22.4LCO2/1molCO2 x 296K/273K = 1.67 L CO2