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Calculate the volume of carbon dioxide produced when 250g of pentane, C5H12 completely combusts at STP. C5H12 + O2 → CO2 + H20 ​

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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.

User Sergey Balashevich
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