Question

Решите задачу по уравнению реакции горения углеводов. Какой объём кислорода необходим для полного сгорания пропана объёмом 44,8 л (н.у.)?

Answer 1

Ответ: 224л кислород

Answer 2

Solving the Combustion Reaction Equation for Propane

To solve the problem of finding the volume of oxygen required for the complete combustion of propane, we can start by writing the balanced chemical equation for the combustion of propane.

The balanced chemical equation for the combustion of propane is as follows: C3H8 + 5O2 → 3CO2 + 4H2O

Now, we can use this balanced equation to determine the volume of oxygen required for the complete combustion of a given volume of propane.

Calculation

Given: Volume of propane (C3H8) = 44.8 L (at standard conditions)

From the balanced chemical equation, we can see that 1 mole of propane reacts with 5 moles of oxygen. Using the molar volume of gases at standard conditions (22.4 L/mol), we can calculate the volume of oxygen required.

1. Calculate the moles of propane: - Moles of propane = Volume of propane / Molar volume of gas at standard conditions - Moles of propane = 44.8 L / 22.4 L/mol - Moles of propane = 2 moles

2. Calculate the moles of oxygen required: - Moles of oxygen = 5 * Moles of propane - Moles of oxygen = 5 * 2 - Moles of oxygen = 10 moles

3. Calculate the volume of oxygen required: - Volume of oxygen = Moles of oxygen * Molar volume of gas at standard conditions - Volume of oxygen = 10 moles * 22.4 L/mol - Volume of oxygen = 224 L

Therefore, the volume of oxygen required for the complete combustion of 44.8 L of propane at standard conditions is 224 liters.