As a result of the combustion of 11.4 grams of alkane, 17.92 liters of CO2 were obtained.

As a result of the combustion of 11.4 grams of alkane, 17.92 liters of CO2 were obtained. Determine which alkane was involved in the reaction.

1. Let’s compose the equation of alkane combustion in general form:
CxH2x + 2 + (1.5x + 0.5) O2 → xCO2 + (x + 1) H2O;
2. Let’s calculate the chemical amount of carbon dioxide produced:
n (CO2) = V (CO2): Vm = 17.92: 22.4 = 0.8 mol;
3. Determine the amount of alkane:
n (CxH2x + 2) = n (CO2): x = 0.8: x mol;
4. Let’s calculate the molecular weight of the unknown alkane:
M (CxH2x + 2) = 12x + 2x + 2 = 14x + 2 g / mol;
M (CxH2x + 2) = m (CxH2x + 2): n (CxH2x + 2) = 11.4: (0.8: x) = 11.4x: 0.8 g / mol;
5. We equate the obtained expressions for the molecular weight and find x:
14x + 2 = 11.4x: 0.8;
11.2x +1.6 = 11.4x;
0.2x = 1.6;
x = 8.
6. Substitute the found value x and write down the alkane formula:
CxH2x + 2 = C8H18.
Answer: C8H18.