Question

Объем кислорода,израсходованного при сгорании 5,6 л метана(н.у)равен: ?

Answer 1

Находим по реакции
 5.6л  х
CH4+2O2=CO2+H2O; х=5.6*44.8/22.4=11.2 литра
22.4л 44.8л

Answer 2

Calculating Oxygen Consumption from Methane Combustion

To calculate the volume of oxygen consumed during the combustion of 5.6 liters of methane at standard conditions, we can use the stoichiometry of the reaction.

The balanced chemical equation for the combustion of methane is: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

From the balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. We can use this information to calculate the volume of oxygen consumed.

Using the ideal gas law, we can convert the volume of methane to moles and then use the mole ratio to find the volume of oxygen consumed.

Calculation

1. Convert the volume of methane to moles: - 5.6 liters of methane at standard conditions is equivalent to 0.2 moles of methane.

2. Use the mole ratio from the balanced equation: - 1 mole of methane reacts with 2 moles of oxygen.

3. Calculate the volume of oxygen consumed: - 0.2 moles of methane will require 0.4 moles of oxygen. - Using the ideal gas law, at standard conditions, 1 mole of gas occupies 22.4 liters. - Therefore, 0.4 moles of oxygen will occupy 8.96 liters.

The volume of oxygen consumed during the combustion of 5.6 liters of methane at standard conditions is approximately 8.96 liters.