How many grams of oxygen are required to react with 10.0 grams of octane (C8H18) in the combustion of octane in gasoline?

Oh fun, stoichiometry. So the equation for this is C8H18+O2----> CO2+H2O, or
2C8H18+25O2----> 16CO2+18H2O when balanced. 10/114.229 g/mol = .087543... mol, converting to moles by dividing by molar mass. The ratio is 25/2, using the integers in front of the molecules in the balanced equation. so you just multiply .087543... by the ratio. That's 1.09429, which you multiply by the molar mass of O2, 31.9988 g/mol. The answer is 35.02 g O2

ardni313 ardni313 · Answer 2 · 2019-07-15T06:08:10+02:00

35.2 grams of oxygen are required to react with 10.0 grams of octane

[tex]\boxed{\boxed{\bold{Further~explanation}}}[/tex]

Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

[tex]\large{\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}[/tex]

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and last atoms O atoms

Octane combustion reaction

C₈H₁₈ + O₂ ---> CO₂ + H₂O

To equalize the reaction equation we give the reaction coefficients with variables a, b and c while the most complex compounds, namely Octane, we give the number 1

So the reaction becomes

C₈H₁₈ + aO₂ ---> bCO₂ + cH₂O

C atom on the left 8, right b, so b = 8

left H atom = 18, right 2c so 2c = 18 ---> c = 9

Atom O on the left 2a, right 2b + 2c, so 2a = 2b + 2c

2a = 2.8 + 9

2a = 16 + 9

2a = 25

a = 25/2

The equation becomes:

C₈H₁₈ + 25/2O₂ ---> 8CO₂ + 9H₂O or be

2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O

To find the mass O2, we find the mole first from the mole ratio with Octane

1. We are looking for the octane mole

Mr. octane = 8.Ar C + 18.Ar H

Mr. octane = 8.12 + 18.1

Mr octane = 96 + 18

Mr octane = 114

known octane mass: 10 grams then the mole:

mol = gram / Mr

mol = 10/ 114

mol = 0.088

2. We look for the O₂ mole

because of the ratio of the reaction coefficient between O2 and octane = 25: 2 then the mole of O₂ =

25/2 x 0.088 = 1,1

So that the mass O₂ =

mole. Mr = 1.1 32

mass = 35.2 grams

[tex]\boxed{\boxed{\bold{Learn~more}}}[/tex]

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Keywords: Complete combustion of Hydrocarbons, alkanes, equalize the reaction equation