Question

Обчисліть об`єм водяної пари, що виділиться при згоранні 4л етену?(з оформленням)

Answer 1

Ответ:

4л Х

С2Н4 + 3О2 = 2СО2 + 2Н2О

22,4л 36г

Х=6,42г=6,42мл

Answer 2

Calculating the Volume of Water Vapor Produced from the Combustion of 4 Liters of Ethene

To calculate the volume of water vapor produced from the combustion of 4 liters of ethene, we can use the balanced chemical equation for the combustion of ethene and apply the principles of stoichiometry.

The balanced chemical equation for the combustion of ethene (C2H4) is as follows: C2H4 + 3O2 → 2CO2 + 2H2O

According to this equation, 1 mole of ethene reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 2 moles of water.

Given that we have 4 liters of ethene, we need to convert this volume to moles and then use the stoichiometric coefficients from the balanced equation to determine the volume of water vapor produced.

First, we need to convert the volume of ethene from liters to moles using the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.

Calculation Steps:

1. Convert 4 liters of ethene to moles: - 4 liters × (1 mole / 22.4 liters) = 0.1786 moles of ethene

2. Use stoichiometry to find moles of water produced: - From the balanced equation, 1 mole of ethene produces 2 moles of water. - Therefore, 0.1786 moles of ethene will produce 0.1786 moles × 2 = 0.3572 moles of water.

3. Convert moles of water to volume at STP: - 0.3572 moles × 22.4 liters/mole = 8.00 liters of water vapor produced.

Conclusion:

The combustion of 4 liters of ethene will produce approximately 8.00 liters of water vapor.