Question

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 ∘C, what is the volume of carbon dioxide formed by the combustion of 2.80 g of butane? what is the volume of CO2?The ideal gas law

PV=nRT

relates pressure P, volume V, temperature T, and number of moles of a gas, n. The gas constant R equals 0.08206 L⋅atm/(K⋅mol)or 8.3145 J/(K⋅mol). The equation can be rearranged as follows to solve for n:

n=PVRT

This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios.

Answer 1

2C4H10(g)+13O2(g) ------------------> 8CO2(g)+10H2O(l)

116g                                                       352g

2.80g                                                        ?

116 g butane -----------------------> 352 g CO2

2.80 g butane ---------------------> x g CO2

x = 352 x 2.80 / 116

x = 8.50 g

CO2 mass released = 8.50 g

CO2 moles = mass / molar mass

                   = 8.50 / 44

                  = 0.193

moles = n = 0.193

T = 273 + 23 = 296 K

P = 1 atm

R = 0.0821 L-atm / mol K

ideal gas equation :

P V = n R T

1 x V = 0.193 x 0.0821 x 296

V = 4.69 L

volume of CO2 released = 4.69 L