If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degrees C, what is the mixture of gasses? answer must be in atm

The total pressure of the gas mixture = 2.76 atm

Note: The question is not complete. The complete question is as follow:

If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degree Celsius , what is the pressure of the mixture of gases?

Explanation:

Mass of O₂ gas = 43.1 g, molar mass of O₂ gas = 32.0 g/mol

Number of moles of O₂ gas = 43.1/32.0 = 1.347 moles

Mass of CO₂ gas = 6.8 g, molar mass of CO₂ gas = 44.0 g

Number of moles of CO₂ gas = 6.8/44 = 0.155 moles

Total number of moles of gas mixture, n = (1.347 + 0.155) = 1.502 moles

Volume of gas mixture, V = 13.7 L

Temperature of gas mixture, T = 34 °C = (273.15 + 34) K = 307.15 K

Pressure of gas mixture = ?

Molar gas constant, R = 0.0821 liter·atm/mol·K.

Using the ideal gas equation: PV =nRT

P = nRT/V

P = (1.502 × 0.0821 × 307.15) / 13.7

P = 2.76 atm

Therefore, the total pressure of the gas mixture = 2.76 atm

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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