How many grams of pentane are fully combusted by 82.9 L of oxygen at STP

Question

How many grams of pentane are fully combusted by 82.9 L of oxygen at STP?

C₅H₁₂ + 8O₂ -----> 5CO₂ + 6H₂O

Answer 1

Here is the plan: You have 82.9 L of oxygen at STP. Since the system is at STP you can use the molar volume of any gas at STP of 22.4L/mole to convert to moles of O₂. Next, you can use mole ratio from the coefficients in the balanced equation to convert from moles of O2 to moles of pentane and finally, multiply by molar mass of pentane (use periodic table to calculate) to get grams of pentane. Round answer to correct number of significant digits which would be 3 digits since the measurement given in the problem has 3 digits.

Applying the plan:

82.9 L O2 x (1 mole O₂/22.4 L O₂) x (1 mole C₅H₁₂/8 moles O₂) x (72.15 g C₅H₁₂/1 mole C₅H₁₂)= 33.4 g C₅H₁₂

Answer 2

First calculate the number of moles of oxygen from the ideal gas equation.

n = pV/RT = 1atm x 82.9L/(0.0821L-atm/mol-deg x 273.15K) = 3.70 moles O₂.

From the balanced chemical equation, we know that the number of moles of pentane is 1/8 the number of moles of oxygen or 3.70/8 = 0.463 moles pentane. As there are 72.15 g pentane per mole, 0.463 x 72.15 = 33. g pentane will react.