# 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.