Question

Каков объём воздуха необходимого для сжигания 44,8 л метана?

Answer 1

CH4 + 2O2 = CO2 + 2H2O
n(CH4) = V/Vm = 44,8/22,4 = 2 моль
n(O2) = 2n(CH4) = 4 моль
V(O2) = n*Vm = 4*22,4 = 89,6 моль
Кислорода в воздухе содержится 21%, поэтому:
Vвозд ---- 100%
89,6 ------- 21%
Vвозд = 89,6*100/21 = 426,7 л

Answer 2

Calculating the Volume of Air Needed to Burn 44.8 L of Methane

To calculate the volume of air needed to burn 44.8 L 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 balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen.

To find the volume of air needed, we can use the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 L.

Using this information, we can calculate the volume of air needed to burn 44.8 L of methane.

The calculation involves converting the volume of methane to moles, then using the stoichiometric ratio to find the moles of oxygen needed, and finally converting the moles of oxygen to volume at STP.

Let's perform the calculation:

1. Converting Volume of Methane to Moles: - 44.8 L of methane at STP can be converted to moles using the relationship 1 mole = 22.4 L. - 44.8 L / 22.4 L/mole = 2 moles of methane.

2. Using Stoichiometric Ratio to Find Moles of Oxygen: - From the balanced equation, 1 mole of methane reacts with 2 moles of oxygen. - Therefore, 2 moles of methane will react with 2 * 2 = 4 moles of oxygen.

3. Converting Moles of Oxygen to Volume at STP: - 4 moles of oxygen at STP will occupy 4 * 22.4 L/mole = 89.6 L of oxygen.

So, the volume of air needed to burn 44.8 L of methane is 89.6 L.