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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

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.

  • CHEM -

    i mean... 0.574L didn't work.

  • CHEM -

    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

    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

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