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

Answer:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

α =2

β = 19

γ = 12

δ = 14

53.2moles of O₂

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 + $\frac{n}{4}$)O₂ → xCO₂ + $\frac{n}{2}$H₂O

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

Therefore:

β = x + $\frac{n}{4}$ = 6 + $\frac{14}{4}$ = 6 + 3.5 = 9$\frac{1}{2}$

γ = 6

δ = $\frac{n}{2}$ = $\frac{14}{2}$ = 7

The complete reaction equation is therefore given as:

                   C₆H₁₄ + 9$\frac{1}{2}$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 = $\frac{5.6 x 19}{2}$

                        k = 53.2moles of O₂