Question

какой объем воздуха потребуется для сжигания 5 л пропан-бутановой смеси массовая доля пропана в которой составляет 5 %

Answer 1

Составляем пропорцию 5л - 100%
                                    Xл - 5 %
5*5/100=0,25л = 250 г
n = 250/15=16.7 моль
V=Vm * n = 22.4 * 16.7 = 373.3 л 

Answer 2

Calculating Air Volume Required for Burning Propane-Butane Mixture

To calculate the air volume required for burning a 5 L propane-butane mixture with a mass fraction of 5% propane, we can use the chemical equation for the combustion of propane:

C3H8 + 5O2 → 3CO2 + 4H2O

This equation shows that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).

Calculation Steps

1. Determine the Moles of Propane and Butane: - Since the mass fraction of propane is 5%, we can assume the remaining 95% is butane. - We'll assume the molar mass of butane is approximately 58.12 g/mol and the molar mass of propane is approximately 44.10 g/mol.

2. Calculate the Moles of Propane and Butane: - Moles of propane = (mass of propane / molar mass of propane) - Moles of butane = (mass of butane / molar mass of butane)

3. Determine the Moles of Oxygen Required: - The stoichiometric ratio of propane to oxygen is 1:5, so we'll use this ratio to calculate the moles of oxygen required.

4. Convert Moles of Oxygen to Volume: - We'll use the ideal gas law (PV = nRT) to convert the moles of oxygen to volume at standard temperature and pressure (STP).

Conclusion

By following these steps, we can calculate the volume of air (oxygen) required to burn the given 5 L propane-butane mixture with a 5% mass fraction of propane.