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 1.00 atm and 23 degrees Celcius, what is the volume of carbon dioxide formed by the combustion of 1.80g of butane?

mols butane = grams/molar mass.

Using the coefficients in the balanced equation, convert mols butane to mols CO2.
Use PV = nRT to solve for V of CO2 at the conditions listed in the problem. Remember T must be in kelvin, P in atm and V will be in L.

1.67 L

your all wrong

To find the volume of carbon dioxide formed by the combustion of butane, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)

First, we need to determine the number of moles of butane. We can use the molar mass of butane to do this.

Molar mass of C4H10 = (4 * atomic mass of Carbon) + (10 * atomic mass of Hydrogen)
= (4 * 12.01 g/mol) + (10 * 1.01 g/mol)
= 58.12 g/mol

Number of moles of butane = Mass of butane (in grams) / Molar mass of butane

Number of moles of butane = 1.80 g / 58.12 g/mol
≈ 0.031 moles

According to the balanced equation, 2 moles of butane produces 8 moles of carbon dioxide.

Therefore, the number of moles of carbon dioxide produced = (0.031 moles butane) * (8 moles CO2 / 2 moles butane)
= 0.124 moles CO2

Next, we need to convert the temperature to Kelvin.

Temperature in Kelvin = 23 degrees Celsius + 273.15
= 296.15 K

Now, we can plug these values into the ideal gas law equation:

PV = nRT

V = (nRT) / P
= (0.124 moles * 0.0821 L.atm/mol.K * 296.15 K) / 1.00 atm
≈ 3.02 L

Therefore, the volume of carbon dioxide formed by the combustion of 1.80g of butane at 1.00 atm and 23 degrees Celsius is approximately 3.02 liters.