Question

Какой объем кислорода потребуется для сжигания 50 л газовой смеси (н.у.), содержащей в равных объемных отношениях метан, этан, пропан, бутан?

Answer 1

СН₄+ 2О₂ = СО₂ + 2Н₂О                        (1)

2С₂Н₆ + 7О₂ = 4СО₂ + 6Н₂О                 (2)

С₃Н₈ + 5О₂ = 3СО₂ + 4Н₂О                   (3)

2С₄Н₁₀ + 13О₂ = 8СО₂ + 10Н₂О           (4)


50 л : 4 = 12,5 л
V(O₂) = (12,5 л * 22,4 л/моль * 2 моль) : (22,4 л/моль * 1 моль) = 25 л  (1)
V(O₂) = (12,5 л * 22,4 л/моль * 7 моль) : (22,4 л/моль * 2 моль) = 43,75 л  (2)
V(O₂) = (12,5 л * 22,4 л/моль * 5 моль) : (22,4 л/моль * 1 моль) = 62,5 л  (3)
V(O₂) = (12,5 л * 22,4 л/моль * 13 моль) : (22,4 л/моль * 2 моль) = 81,25 л  (4)
V(O₂ общ) = 25 + 43,75 + 62,5 + 81,25 = 212,5 л

Answer 2

Calculating Oxygen Volume Required for Burning Gas Mixture

To calculate the volume of oxygen required for burning a gas mixture containing methane, ethane, propane, and butane in equal volumetric ratios, we can use the stoichiometric coefficients of the combustion reaction for each gas. The balanced chemical equation for the complete combustion of these hydrocarbons is as follows:

CH4 + 2O2 → CO2 + 2H2O C2H6 + 3.5O2 → 2CO2 + 3H2O C3H8 + 5O2 → 3CO2 + 4H2O C4H10 + 6.5O2 → 4CO2 + 5H2O

The stoichiometric coefficients indicate the molar ratio of each gas to oxygen in the combustion reaction.

Calculating Oxygen Volume

To calculate the oxygen volume required for burning 50 L of the gas mixture, we can use the stoichiometric coefficients and the given equal volumetric ratios of the gases to determine the total oxygen volume required.

- For methane (CH4), the stoichiometric coefficient is 2. - For ethane (C2H6), the stoichiometric coefficient is 3.5. - For propane (C3H8), the stoichiometric coefficient is 5. - For butane (C4H10), the stoichiometric coefficient is 6.5.

Using the given equal volumetric ratios, the total stoichiometric coefficient for the gas mixture can be calculated as follows:

Total Stoichiometric Coefficient = (2 + 3.5 + 5 + 6.5) = 17

Now, to find the volume of oxygen required, we can use the following calculation:

Volume of Oxygen Required = Total Stoichiometric Coefficient * Volume of Gas Mixture Volume of Oxygen Required = 17 * 50 L Volume of Oxygen Required = 850 L

Therefore, the volume of oxygen required for burning 50 L of the gas mixture is 850 liters.