Amajor component of gasoline is octane c8h18 . when octane is burned in air, it chemically reacts with oxygen gas o2 to produce carbon dioxide co2 and water h2o . what mass of oxygen gas is consumed by the reaction of 6.0g of octane?

The reaction involved in the problem is called a combustion reaction. The balanced chemical reaction is expressed as:

C8H18 + 25/2O2  = 8CO2 + 9H2O

TO determine the amount of oxygen gas needed to completely react the given amount of C8H18, we use the amount of the other reactant and the relation of the substances from the chemical equation. We do as follows:

6.0 g C8H18 ( 1 mol C8H18 / 114.26 g C8H18) ( 25/2 mol O2 / 1 mol C8H18 ) ( 32 g O2 / 1 mol O2 ) = 21.00 g O2

Therefore, to completely react 6 g of octane we need at least 21 g of oxygen gas.

The mass of oxygen gas consumed by the reaction of 6.0 g of octane is  21 .0 g

Further Explanation  Hydrocarbons  Hydrocarbons are types of compounds that are mostly made up of hydrogen and carbon elements. However other hydrocarbons are made up of carbon, hydrogen and oxygen elements.The major types of hydrocarbons include, alkanes, alkenes, alkynes, alcohols, and alkanoic acids.  Alkanes are saturated hydrocarbons while alkenes and alkynes are unsaturated hydrocarbons containing only carbon and hydrogen atoms.  Alcohols and alkanoic acids contains hydrogen, oxygen and carbon atoms.Combustion of Hydrocarbons  Hydrocarbons burn in air to form carbon dioxide and water as the only products.

That is;

For hydrocarbons with carbon and hydrogen

CxHy + O2(g) → CO2(g) + H2O(g)  

For Example;

Combustion of octane;

2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(l)

Therefore; Two moles of octane combines with 25 moles of oxygen gas to yield 16 moles of carbon dioxide and 18 moles of water.

Therefore, we can calculate the mass of oxygen consumed by the reaction of 6.0 g of octane.

Step 1; Number of moles of octane

Number of moles = mass/ molar mass

Molar mass of octane = 114.23 g/mol

Number of moles of octane in 6.0 g

            = 6.0 g/114.23 g/mol

            = 0.0525 moles

Step 2: Number of moles of Oxygen gas used

From the balanced equation;

Two moles of octane combines with 25 moles of oxygen gas to yield 16 moles of carbon dioxide and 18 moles of water.

Therefore; moles of oxygen gas = (0.0525/2)× 25

                                                     = 0.65625 moles of Oxygen gas

Step 3: Mass of oxygen gas

Mass = number of moles × molar mass

Molar mass of oxygen gas = 32.0 g/mol

Mass of oxygen = 0.65625 moles × 32.0 g/mole

                          = 21 g of oxygen gas

Keywords: Combustion, hydrocarbons  

Learn more about:  Hydrocarbons: Combustion of hydrocarbons: Major types of hydrocarbons:

Level: High school

Subject: Chemistry  

Topic: Organic chemistry  

Sub-topic: Combustion of hydrocarbons

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