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 (0 degrees C, 1 Atm)?

C₅H₁₂ + 8O₂ ----> 5CO₂ + 6H₂O

Answer 1

Like any stoichiometry problem, if we want to compare quantities of different reactants or products, we must do so in moles. Therefore, we can start by finding the number of moles of O₂ combusted by using the ideal gas law. Don’t forget to use Kelvin temperatures!

PV = nRT; (1 atm)(82.9 L) = (n)(0.0821 L*atm/K*mol)(273 K); n = 3.70 mol O₂

With the number of moles of O₂ in hand, we can first convert to moles of pentane using our mole-to-mole ratio, then to grams of pentane using its molar mass.

3.70 mol O₂ * 1 mol C₅H₁₂/8 mol O₂ * 72.151 g C₅H₁₂/1 mol C₅H₁₂ = 33.4 g C₅H₁₂

I hope this helped!