Answer: 88.02g of CO2
Explanation:
Find the molar mass of ethane (C2H6):
Molar mass of C2H6 = 2(12.01 g/mol) + 6(1.01 g/mol) = 24.02 g/mol
Calculate the number of moles of ethane:
Moles of C2H6 = Mass (g) / Molar mass (g/mol)
Moles of C2H6 = 24.0 g / 24.02 g/mol ≈ 1.000 moles
Since 1 mole of ethane produces 2 moles of carbon dioxide (CO2), you'll have 2 moles of CO2 for every mole of ethane combusted.
Calculate the moles of carbon dioxide produced:
Moles of CO2 = 2 moles of CO2 per mole of C2H6 * Moles of C2H6
Moles of CO2 = 2 moles/mole * 1.000 moles ≈ 2.000 moles
Finally, calculate the mass of carbon dioxide produced:
Mass of CO2 = Moles of CO2 * Molar mass of CO2
Mass of CO2 = 2.000 moles * (12.01 g/mol + 2 * 16.00 g/mol) ≈ 88.02 g
So, approximately 88.02 grams of carbon dioxide will be produced when 24.0 grams of ethane is combusted in the presence of excess oxygen.