Final answer:
The volume of carbon dioxide produced from the complete combustion of 250g of pentane at STP is calculated using stoichiometry and is found to be 388.571 liters.
Step-by-step explanation:
The volume of carbon dioxide produced from the combustion of pentane (C5H12) can be calculated using stoichiometry and the molar volume of a gas at Standard Temperature and Pressure (STP). First, the balanced equation for the complete combustion of pentane is:
C5H12 + 8 O2 → 5 CO2 + 6 H2O
Next, we calculate the number of moles of pentane:
Molar mass of C5H12 = (5 × 12.01) + (12 × 1.008) = 72.15 g/mol
Moles of C5H12 = 250 g ÷ 72.15 g/mol = 3.465 moles
The stoichiometry of the balanced equation indicates that for every mole of pentane, five moles of carbon dioxide are produced. Therefore:
Moles of CO2 = 3.465 moles of C5H12 × 5 moles CO2/mole C5H12 = 17.325 moles CO2
At STP, 1 mole of a gas occupies 22.414 liters. Thus, the volume of CO2 produced is:
Volume of CO2 = 17.325 moles × 22.414 L/mole = 388.571 L
This is the volume of carbon dioxide produced at STP from the complete combustion of 250g of pentane.