C2H6 + O2 -> CO2 + H2O How many g of carbon dioxide are produced at the same time as 360g of water vapor in the burning of ethane

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587 To determine the mass of carbon dioxide produced in the burning of ethane, we need to calculate the mole ratio between carbon dioxide and water vapor. From the balanced equation:

C2H6 + O2 -> CO2 + H2O

we can see that the mole ratio between CO2 and H2O is 1:1 (1 mole of CO2 is produced for every 1 mole of H2O).

To calculate the number of moles of water vapor, we need to convert the given mass of water vapor (360g) into moles. We can do this by using the molar mass of water (H2O), which is approximately 18.015 g/mol.

Number of moles of H2O = Mass of H2O / Molar mass of H2O
= 360g / 18.015 g/mol
= 19.98 mol (approximately)

Since the mole ratio between CO2 and H2O is 1:1, the number of moles of carbon dioxide produced is also 19.98 mol.

To calculate the mass of carbon dioxide produced, we multiply the number of moles of CO2 by its molar mass. The molar mass of carbon dioxide (CO2) is approximately 44.01 g/mol.

Mass of CO2 = Number of moles of CO2 * Molar mass of CO2
= 19.98 mol * 44.01 g/mol
= 879.94 g (approximately)

Therefore, approximately 880 grams of carbon dioxide are produced simultaneously with 360 grams of water vapor during the burning of ethane.