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.
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
To calculate the volume of carbon dioxide produced by the combustion of 1.00g of butane, you need to use the ideal gas law equation:
PV = nRT
where:
P is the pressure, which is given as 1.00 atm
V is the volume
n is the number of moles of the gas
R is the ideal gas constant (0.082058 L atm/mol K)
T is the temperature in Kelvin (296.15 K)
First, let's calculate the number of moles of CO2 produced:
0.017204 mol C4H10 x (8 mol CO2 / 2 mol C4H10) = 0.068816 mol CO2
Now we can substitute the values into the ideal gas law equation to find the volume:
(1.00 atm)(V) = (0.068816 mol)(0.082058 L atm/mol K)(296.15 K)
V = (0.068816 mol)(0.082058 L atm/mol K)(296.15 K) / (1.00 atm)
V ≈ 1.77 L
So, approximately 1.77 liters of carbon dioxide are formed by the combustion of 1.00g of butane at 1.00 atm and 23°C.
To find the volume of carbon dioxide formed by the combustion of 1.00g of butane at 1.00atm and 23°C, you need to use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P is the pressure (in atm),
V is the volume (in liters),
n is the number of moles of gas,
R is the ideal gas constant (0.082058 L atm/mol K), and
T is the temperature (in Kelvin).
First, let's calculate the number of moles of carbon dioxide produced from 1.00g of butane:
1.00g of butane = 0.017204 mol C4H10 (which you correctly calculated)
Now, using the balanced equation of the combustion reaction, we can see that for every 2 moles of butane combusted, 8 moles of carbon dioxide are produced. Therefore, the number of moles of carbon dioxide formed is:
0.017204 mol C4H10 * (8 mol CO2 / 2 mol C4H10) = 0.068816 mol CO2
Next, convert the temperature from Celsius to Kelvin:
23°C + 273.15 = 296.15K
Now, you can plug the values into the ideal gas law equation:
PV = nRT
(1.00atm) * V = (0.068816 mol) * (0.082058 L atm/mol K) * (296.15K)
Solving for V:
V = (0.068816 mol * 0.082058 L atm/mol K * 296.15K) / 1.00atm
V = 1.5996 L (rounded to four decimal places)
Therefore, the volume of carbon dioxide formed by the combustion of 1.00g of butane at 1.00atm and 23°C is 1.5996 liters.