A mixture of carbon dioxide and carbon monoxide occupies a volume of 15 liters at a temperature

A mixture of carbon dioxide and carbon monoxide occupies a volume of 15 liters at a temperature of 25 degrees and a pressure of 730 mm Hg. Art. The mass of this mixture is 17.5g. Determine the volume fraction of carbon dioxide

1.for gases under conditions different from normal, the following equation is valid:

P * V = n * R * T;

n = (P * V): (R * T);

2. bring the temperature to standard units:

T = 25 + 273 = 298 K;

3. Let’s translate the pressure from mm Hg. Art. in kPa:

760 mmHg Art. = 101.325 kPa;

730 mmHg st = z;

z = (730 * 101.325): 760 = 97.325 kPa;

4. calculate the chemical amount of the mixture:

n (mixture) = (97.325 * 15): (8.314 * 298) = 0.589 mol;

5. find the molecular weight of the mixture:

M (mixture) = m (mixture): n (mixture);

M (mixture) = 17.5: 0.589 = 29.7 g / mol;

6.To calculate the molecular weight of the mixture, use the formula with volume fractions:

M (mixture) = ϕ (CO2) * M (CO2) + ϕ (CO) * M (CO);

M (CO2) = 12 + 2 * 16 = 44 g / mol;

M (CO) = 12 + 16 = 28 g / mol;

7. ϕ (СО2) + ϕ (СО) = 1

let ϕ (СО2) = х, then ϕ (СО) = 1 – х;

8.we substitute the expressions for the volume fractions in the formula for finding the molecular weight of the mixture:

M (mixture) = x * 44 + (1 – x) * 28 = 29.7;

9.find x:

44 * x + 28 – 28 x = 29.7;

16 * x = 29.7 – 28;

16 * x = 1.7;

x = 0.10625, that is, ϕ (CO2) = 0.10625 or 10.625%.

Answer: 10.625% CO2.