Propane burns in air according to the equation: C₃H₈(g) + 5O₂(g) $\rightarrow$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

We can use the mole ratio between propane and oxygen from the balanced chemical equation to calculate the volume of oxygen required.

1 mol of propane reacts with 5 mol of oxygen.

We can use the ideal gas law to relate the volume of a gas to its number of moles, pressure, temperature, and gas constant. Since the gases are under the same conditions, the pressure, temperature, and gas constant are constant.

PV = nRT

Assuming standard temperature and pressure (STP), the gas constant (R) is 0.0821 L·atm/mol·K, and the temperature (T) is 273 K. Thus, we can simplify the ideal gas law to:

V = n

We can use the formula:

n = PV/RT

to calculate the number of moles of propane (n) from its volume (V) and assuming it is at STP.

n = (8.00 L)(1 atm)/(0.0821 L·atm/mol·K)(273 K)

n = 0.314 mol propane

Since 1 mol of propane reacts with 5 mol of oxygen, we need:

n = 5(0.314 mol) = 1.57 mol oxygen

Finally, we can calculate the volume of oxygen required:

V = n = (1.57 mol)(22.4 L/mol) = 35.1 L

Therefore, the best choice is B) 40.0 L.

Propane Burns in Air According to the Equation: C3H8(g) →\rightarrow→ 3CO2(g) + 4H2O(g) What Volume of O2 would Be Required

Propane Burns in Air According to the Equation: C₃H₈(g) $\rightarrow$ 3CO₂(g) + 4H₂O(g) What Volume of O₂would Be Required