How many grams of oxygen are required to react with 16.0 grams of octane (c8h18) in the combustion of octane in gasoline?

Answer is: 56 grams of oxygen.
Chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ +9H₂O or
2C₈H₁₈ + 25O₂ → 16CO₂ +18H₂O.
m(C₈H₁₈) = 16 g.
m(O₂) = ?
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈)
n(C₈H₁₈) = 16 g ÷ 114,2 g/mol.
n(C₈H₁₈) = 0,14mol.
from reaction: n(C₈H₁₈) : n(O₂) = 2 : 25.
n(O₂) = 1,75 mol.
m(O₂) = 1,75 mol · 32 g
m(O₂) = 56 g.

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ardni313 ardni313 · Answer 2 · 2019-07-14T01:26:29+02:00

56 grams of oxygen are required to react with 16.0 grams of octane C₈H₁₈

C₈H₁₈ including alkane hydrocarbons which when burned perfectly will produce CO₂ and H₂O

[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 H₂O

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

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: 16 grams then the mole:

mole = gram / Mr

mole = 16/ 114

mole = 0.14

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 0.14 = 1.75

So that the mass O₂ =

mole. Mr = 1.75 32

mass = 56 grams

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

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