Question

Consider the following unbalanced equation for the combustion of hexane: αC6H14(g)+βO2(g)→γCO2(g)+δH2O(g) Part A Balance the equation. Give your answer as an ordered set of numbers α, β, γ, ... Use the least possible integers for the coefficients. α α , β, γ, δ = nothing Request Answer Part B Determine how many moles of O2 are required to react completely with 5.6 moles C6H14. Express your answer using two significant figures. n n = nothing mol Request Answer Provide Feedback

Answer 1

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

α =2

β = 19

γ = 12

δ = 14

53.2moles of O₂

Step-by-step explanation:

Proper equation of the reaction:

αC₆H₁₄ + βO₂ → γCO₂ + δH₂O

This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:

CₓHₙ + (x +
`(n)/(4)`)O₂ → xCO₂ +
`(n)/(2)`H₂O

From the given combustion equation, x = 6 and n = 14

Therefore:

β = x +
`(n)/(4)` = 6 +
`(14)/(4)` = 6 + 3.5 = 9
`(1)/(2)`

γ = 6

δ =
`(n)/(2)` =
`(14)/(2)` = 7

The complete reaction equation is therefore given as:

C₆H₁₄ + 9
`(1)/(2)`O₂ → 6CO₂ + 7H₂O

To express as whole number integers, we multiply the coefficients through by 2:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

Problem 2

From the reaction:

2 moles of hexane are required to completely react with 19 moles of O₂

∴ 5.6 moles of hexane would react with k moles of O₂

This gives: 5.6 x 19 = 2k

k =
`(5.6 x 19)/(2)`

k = 53.2moles of O₂