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)+13O(g)-> 8CO2(g)+10H2O(l)

At 1.00 atm and 23C , what is the volume of carbon dioxide formed by the combustion of 3.40 grams of butane?

PV=nRT or n=PV/RT
R=0.08206

To find the volume of carbon dioxide formed by the combustion of 3.40 grams of butane, we need to calculate the number of moles of butane and use the balanced equation to determine the number of moles of carbon dioxide produced. Then we can use the ideal gas law to find the volume of carbon dioxide.

Step 1: Calculate the number of moles of butane.
To calculate the number of moles of butane, we need to use the molar mass of butane (C4H10), which is the sum of the molar masses of carbon (C) and hydrogen (H).

Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol

Molar mass of C4H10 = 4(12.01 g/mol) + 10(1.008 g/mol)

Step 2: Calculate the number of moles of butane.
To calculate the number of moles of butane, we can use the formula:

moles = mass / molar mass

moles of butane = 3.40 g / molar mass of C4H10

Step 3: Use the balanced equation to determine the number of moles of carbon dioxide.
From the balanced equation, we can see that for every 2 moles of butane combusted, 8 moles of carbon dioxide are formed.

moles of carbon dioxide = (moles of butane) x (8 moles of CO2 / 2 moles of C4H10)

Step 4: Use the ideal gas law to find the volume of carbon dioxide.
We can use the ideal gas law equation, PV = nRT, to find the volume of carbon dioxide.

V = (n x R x T) / P

where:
V = volume of carbon dioxide (unknown)
n = number of moles of carbon dioxide
R = ideal gas constant (0.08206 L·atm·mol^−1·K^−1)
T = temperature in Kelvin (23°C = 23 + 273 = 296 K)
P = pressure (1.00 atm)

V = (moles of carbon dioxide x R x T) / P

Now you can plug in the values to calculate the volume of carbon dioxide formed by the combustion of 3.40 grams of butane.