Question

What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline? 2C8H18 + 25O2 ⟶ 16CO2 + 18H2O

Answer 1

[AI-generated answer — Bot GPT 3.5]

To determine the mass of oxygen gas consumed in the combustion of 702 g of octane, we first need to find the balanced chemical equation and use stoichiometry to calculate the amount of oxygen consumed.

From the balanced chemical equation given:
2 moles of octane (C8H18) react with 25 moles of oxygen gas (O2) to produce 16 moles of carbon dioxide (CO2) and 18 moles of water (H2O).

1 mole of octane (C8H18) has a molar mass of 114.23 g
702 g of octane is equivalent to 702 / 114.23 = 6.14 moles of octane

According to the balanced equation, 2 moles of octane react with 25 moles of oxygen gas.
Therefore, 6.14 moles of octane will react with (6.14 x 25 / 2) = 76.75 moles of oxygen gas.

The molar mass of oxygen gas (O2) is 32.00 g/mol.
Therefore, the mass of oxygen gas consumed will be 76.75 x 32 = 2,456 g.

Thus, the mass of oxygen gas consumed in the combustion of 702 g of octane is 2,456 g.