Propane burns in air according to the equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
What volume of O₂would be required 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, 5 moles of oxygen gas are needed to react with 1 mole of propane. Therefore, the number of moles of oxygen gas needed to react with 1 mole of propane is (5/1) = 5. To determine the number of moles of propane in 8.00 L of propane, we need to use the ideal gas law: PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature. Assuming that all the gases are under the same conditions, we can write:

n = (PV)/(RT)

At standard temperature and pressure (STP), which is 0°C and 1 atm, the molar volume of an ideal gas is 22.4 L/mol. Therefore,

n = (8.00 L)/(22.4 L/mol) = 0.357 mol

This means that 0.357 mol of propane reacted with 5 times as many moles of oxygen gas. Therefore, the number of moles of oxygen gas needed to react with 8.00 L of propane is:

(5 mol O2) x (0.357 mol propane) = 1.79 mol O2

To convert this to volume at STP, we use the molar volume again:

V = (n)(22.4 L/mol) = (1.79 mol)(22.4 L/mol) = 40.0 L

Therefore, the answer is B) 40.0 L.

Propane Burns in Air According to the Equation: C3H8(g)+ 5O2(g)→

Propane Burns in Air According to the Equation: C₃H₈(g)+ 5O₂(g)→