Suppose 0.410 kg of hexane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane

Explanation:

We first write a balanced reaction for the complete combustion of hexane thus

The stoichiometry of the cumbustion of hexane in air is

2C6H14 (g) + 18O2 (g) →12CO2 (g) + 14H2O (l) or

C6H14 (g) + 9O2 (g) →6CO2 (g) + 7H2O (l)

From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide

Therefore we calculate the nuber of moles of hexane present in the sample thus

Mass hexane of sample = 0.41 kg

Molar nass of hexane = 86.18 g/mol

number of moles of hexane = (mass of hexane) / (molar mass of hexane) = (0.41*1000) / 86.16 = 410/86.16 = 4.76 moles

As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces

4.76*6 moles of CO2 which is 28.55 moles of CO2

From the question we have the temperature and the pressure of the production of CO2 as

Temperature of reaction = 13° C converting to kelving gives = 13+273.15 = 286.15 K

and pressure = 1 atmosphere or 101325 Pa

13.0∘C=13.0∘C+273.15=286.15 K

The volume of the produced CO2 can be calculated using the combined ideal gas equation given by

P*V=n*R*T where

Here

P = Gas pressure (of CO2)

V = Volume (of the CO2)

n = number of moles of gas (CO2) present

R = universal gas constant, equal to 0.0821 atm * L / (mol * K)

T = absolute temperature in Kelvin

Thus we have

1*V = 28.55*0.0821*286.15 or V = 670.76L

Rounding up the answer to 3 significant digits we have

670.76L ≅ 671L

671 litres of carbon dioxide is produced from 0.410 kg of hexane