Question

A mixture of hydrocarbons contains 38.3% hexane, C6H14, 13.9% octane, C8H18, and 47.8% decane, C10H22. The mixture is combusted in an excess of oxygen. What mass of CO2 is generated from the combustion of 19.3 kg of the mixture

Answer 1

Answer:

52.206 kg

Explanation:

From the given information:

Mass of hexane C6H14 = $19.3*10^3 \ g \times \dfrac{38.3}{100}$

= 7391.9 g

Mass of octane C8H18 = $19.3*10^3 \ g \times \dfrac{13.9}{100}$

= 2682.7 g

Mass of decane C10H22 = $19.3*10^3 \ g \times \dfrac{47.8}{100}$

= 9225.4 g

However, recall that:

number of moles of an atom = mass/molar mass

∴

For hexane, no of moles = 7391.9 g/86.18 g/mol

= 85.77 moles

For octane, no of moles = 2682.7 g/114.23 g/mol

= 23.49 moles

For decane, no of moles = 9225.4 g/142.29 g/mol

= 64.84 moles

Therefore:

number of moles of CO2 produced = (6 × 85.77)+(23.49)+(10×64.84) moles

= 1186.51 moles

Finally, the mass of CO2 produced is:

= 1186.51 mol × 44 g/mol

= 52206.44  g

= 52.206 kg