Question

Сколько литров воздуха (н.у.) потребуется для сжигания 96г метана? Объемная доля кислорода в воздухе 21%

Answer 1

96г___ Хл
CH4+2O2=CO2+2H2O
16г___2×22,4л

х=96г×44,8л/16г= 268.8л

О2 в воздухе 21%
21 % = 268.8л
100% = Хл

х=100%× 268.8л/21%= 1 280л
ответ для сжигания 96г метана понадобится 1 280л воздуха

Answer 2

Calculating Air (Dry) Requirement for Combustion of Methane

To calculate the amount of air (dry) required for the combustion of 96g of methane, we can use the stoichiometric ratio of methane to oxygen in the combustion reaction.

The balanced chemical equation for the combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2).

Calculation Steps

1. Convert grams of methane to moles: - The molar mass of methane (CH4) is approximately 16g/mol. - Therefore, 96g of methane is equal to 96g / 16g/mol = 6 moles of methane.

2. Determine the moles of oxygen required: - According to the stoichiometry of the balanced equation, 1 mole of methane requires 2 moles of oxygen. - Therefore, 6 moles of methane will require 6 moles * 2 = 12 moles of oxygen.

3. Calculate the volume of air (dry) required: - The volume of air (dry) can be calculated using the ideal gas law, where 1 mole of gas occupies 22.4 liters at standard temperature and pressure (STP).

Answer

The volume of air (dry) required for the combustion of 96g of methane is approximately 268.8 liters at STP.

Cited Sources

- Source ' .pdf' - Source ' ' - Source ' '