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Finger [1]
3 years ago
8

Determine the enthalpy of the products of ideal combustion, i.e., no dissociation, resulting from the combustion of an isooctane

-air mixture
Chemistry
1 answer:
jeka943 years ago
8 0
The combustion of isooctane (C₈H₁₈) is written below:

2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)

The formula for heat of combustion is:

ΔHc = (∑Stoichiometric coefficient×ΔHf of products) - (∑Stoichiometric coefficient×ΔHf of reactants), where ΔHf is heat of formation.

ΔHf of isooctane = -259.2 kJ/mol
ΔHf of O₂ = 0 kJ/mol
ΔHf of CO₂ = -393.5 kJ/mol 
ΔHf of H₂O = <span>-285.8 kJ/mol
</span>
ΔHc = [(16 mol×-393.5 kJ/mol )+(18 mol×-285.8 kJ/mol)] - [(2 mol×-259.2 kJ/mol) + (25 mol*0 kJ/mol)]
ΔHc = -10,922 kJ
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The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of
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Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms  of C , H , O in the compound

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= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

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= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

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