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) = 0.017204molC4H10
Then: mol CO2= 1g*1mol/44.011g = 0.2272mol
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 .0547L as an answer but it wasn't correct.
2 answers
i mean... 0.574L didn't work.
I think I would convert from 1 mole of a gas occupying 22.4 litres at 273.15 K and 1 atm (STP).
So calculate the volume of CO2 at STP (22.4 litres x number of moles of CO2)
then convert to 296.15 K using
V1/T1=V2/T2
A comment on your calculation of moles:
I agree with your number of moles of butane as 0.0172 moles
but there must be 4 x the number of moles of CO2 as each mole of butane gives 4 moles fo CO2.
4 x 0.0172 moles = 0.0688 moles of CO2
So calculate the volume of CO2 at STP (22.4 litres x number of moles of CO2)
then convert to 296.15 K using
V1/T1=V2/T2
A comment on your calculation of moles:
I agree with your number of moles of butane as 0.0172 moles
but there must be 4 x the number of moles of CO2 as each mole of butane gives 4 moles fo CO2.
4 x 0.0172 moles = 0.0688 moles of CO2