Question

Какой объём кислорода, при нормальных условиях, необходим для полного сжигания 5 м³ метана CH₄

Answer 1

 5м3   V м3

CH4+2O2=CO2+2H2O
1 М3  2 М3
5/1=V/2
V=5*2/1=10 м3 O2

Answer 2

5*103л  V-?
CH4    +2O2   =CO2+2H2O  5m3=5*103л
22,4л     44,8л
V(O2)= 5*103*44,8/22,4=10*103(л)=10м3
Ответ: 10м3

Answer 3

Oxygen Volume Required for Complete Combustion of Methane

To calculate the volume of oxygen required for the complete combustion of 5 m³ of methane (CH₄) under normal conditions, we can use the balanced chemical equation for the combustion of methane.

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

From this equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

Using the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 L/mol, we can calculate the volume of oxygen required.

Given: - Volume of methane (CH₄) = 5 m³

Calculation:

1. Calculate the moles of methane: - Moles of CH₄ = Volume of CH₄ / Molar volume at STP - Moles of CH₄ = 5 m³ * (1000 L / 1 m³) / 22.4 L/mol - Moles of CH₄ ≈ 223.21 moles

2. Calculate the moles of oxygen required: - Moles of O₂ = Moles of CH₄ * (2 moles of O₂ / 1 mole of CH₄) - Moles of O₂ ≈ 223.21 moles * 2 - Moles of O₂ ≈ 446.42 moles