Propane burns in air according to the equation: C₃H₈(g) + 5O₂(g) $\rightarrow$ 3CO₂(g) + 4H₂O(g) What volume of CO₂would be formed if 8.00 L of propane burns, assuming that all of the gases are under the same conditions?

Question

Quizplus · Accepted Answer

According to the balanced equation, for every 1 mole of propane burned, 3 moles of CO2 are produced. Therefore, we need to determine the number of moles of propane present in 8.00 L.

Using the ideal gas law, PV = nRT, we can rearrange to solve for n:

n = PV/RT

Assuming standard temperature and pressure (STP), which is 273 K and 1 atm, we can convert the volume to moles:

n = (1 atm)(8.00 L)/(0.0821 L*atm/mol*K)(273 K) = 0.317 mol propane

Since the ratio of propane to CO2 is 1:3, we can multiply the moles of propane by 3 to find the moles of CO2 produced:

0.317 mol propane x 3 = 0.951 mol CO2

Finally, we can convert the moles of CO2 to volume using the ideal gas law:

V = nRT/P

V = (0.951 mol)(0.0821 L*atm/mol*K)(273 K)/(1 atm) = 23.6 L

Therefore, the volume of CO2 produced is 24.0 L, which is choice C.

Propane Burns in Air According to the Equation: C3H8(g) →\rightarrow→ 3CO2(g) + 4H2O(g) What Volume of CO2 would Be Formed

Propane Burns in Air According to the Equation: C₃H₈(g) $\rightarrow$ 3CO₂(g) + 4H₂O(g) What Volume of CO₂would Be Formed