Question

Рассчитайте массу кислорода необходимого для получения полного сжигания 30 литров азота

Answer 1

N2+2O2=2NO2
n=V/Vm
n(N2)=30/22,4=1,34mol
n(O2)=n(N2)*2
n(O2)=1,34*2=2,68mol
m=n*M
m(O2)=2,68*32g/mol=85,7g

Answer 2

Calculating the Mass of Oxygen Required for Complete Combustion of 30 Liters of Nitrogen

To calculate the mass of oxygen required for the complete combustion of 30 liters of nitrogen, we can use the balanced chemical equation for the combustion of nitrogen:

2N2 + 5O2 → 2N2O5

From the balanced chemical equation, we can see that 2 moles of nitrogen (N2) react with 5 moles of oxygen (O2) to produce 2 moles of dinitrogen pentoxide (N2O5).

Given that the volume of the gas is 30 liters, we need to convert the volume of nitrogen to moles using the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.

Using the ideal gas law, we can calculate the number of moles of nitrogen: ``` n = V / V_m n = 30 L / 22.4 L/mol n ≈ 1.34 mol ``` Now that we have the number of moles of nitrogen, we can use the stoichiometry of the balanced chemical equation to determine the amount of oxygen required.

According to the balanced chemical equation, 2 moles of nitrogen react with 5 moles of oxygen. Therefore, the number of moles of oxygen required can be calculated as: ``` moles of oxygen = (5/2) * moles of nitrogen moles of oxygen = (5/2) * 1.34 moles of oxygen ≈ 3.35 mol ``` Finally, to calculate the mass of oxygen required, we can use the molar mass of oxygen, which is approximately 16 g/mol: ``` mass of oxygen = moles of oxygen * molar mass of oxygen mass of oxygen ≈ 3.35 mol * 16 g/mol mass of oxygen ≈ 53.6 g ```

Therefore, the mass of oxygen required for the complete combustion of 30 liters of nitrogen is approximately 53.6 grams.

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