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αc6h14(g)+βo2(g)→γco2(g)+δh2o(g) part a balance the

source : study-assistant.com

αc6h14(g)+βo2(g)→γco2(g)+δh2o(g) part a balance the

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 + )O₂ → xCO₂ + H₂O

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

Therefore:

β = x + = 6 + = 6 + 3.5 = 9

γ = 6

δ = = = 7

The complete reaction equation is therefore given as:

                   C₆H₁₄ + 9O₂ → 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 =

                        k = 53.2moles of O₂

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