Gas Mixture Composition: Density, Partial Pressure, and Combustion Volume

a) What is the density of this gas at 25°C and 755 mmHg, in grams per liter?

Final answer: The density of the gas mixture is 1.235 g/L. Choose the correct option.

A) 0.865 g/L

B) 1.025 g/L

C) 1.125 g/L

D) 1.235 g/L

b) What is the partial pressure of CO in this mixture at 0.00°C and 1 atm?

Final answer: The partial pressure of CO in the mixture is 0.22 atm. Choose the correct option.

A) 0.22 atm.

B) 0.44 atm.

C) 0.88 atm.

D) 1.00 atm.

c) What volume of air is required for the complete combustion of this mixture of gas?

Final answer: The volume of air required for complete combustion is 643.53 L. Choose the correct option.

A) 107.18 L.

B) 204.70 L.

C) 490.12 L.

D) 643.53 L.

Density of Gas Mixture, Partial Pressure of CO, and Combustion Volume

To calculate the density of a gas mixture, we need to use the ideal gas law, PV = nRT. Given the composition of the gas mixture and the conditions of temperature and pressure, we can calculate the density of the gas mixture to be 1.235 g/L, which corresponds to option D.

To find the partial pressure of CO in the mixture, the partial pressure formula, P = (Patm) X (percent content in mixture) can be used. The partial pressure of CO in the mixture is 0.22 atm, which corresponds to option A.

To determine the volume of air required for the complete combustion of the gas mixture, we need to consider the stoichiometry of the combustion reaction. By comparing the moles of oxygen in the gas mixture to the moles of oxygen required for complete combustion, we can calculate the volume of air needed, which is 643.53 L, corresponding to option D.