Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:
(1) Complete combustion forms CO2 and water vapor.
(2) Incomplete combustion forms CO and water vapor.
(3) Some wax is oxidized to elemental C (soot) and water vapor.
a. find delta H combustion of each reaction for 1 mol C21H44 (delta Hf of C21H44 = -476 kJ/mol; use graphite for elemental carbon)
b. find q (in kJ) when a 413-g candle burns completely
c. find q (in kJ) when 5.19% mass of the candle burns incompletely and another 6.50% undergoes soot formation

(a)

-1.31 x 10⁴ kJ/mol complete reaction

-7.17 x 10³ kJ/mol incomplete reaction

-4.84 x 10³ kJ/mol soot formation

(b) -1.82 x 10⁴ kJ

(c) -1.70 x 10⁴ kJ

Explanation:

We need to find the change in enthalpy  ( which is equivalent to q ) for the following reactions:

(1) C₂₁H₄₄ (s) + 32 O₂ (g) →  21 CO₂ (g) + 22 H₂O (g)    

(2) C₂₁H₄₄ (s) + 43/2 O₂ (g) → 21 CO(g) + 22 H₂O(g)

(3) C₂₁H₄₄ (s) + 11 O₂(g)  →  21 C(graphite) + 22 H₂O(g)

The enthalpy change for a reaction can be calculated from the heats of formation of reactants and products.

We were only given the heat of formation for C₂₁H₄₄, so we will need to find the rest.

ΔHºf (O₂) = ΔHºf(graphite) = 0 (They are in their elemental form)

ΔHºf (H₂O) = -241.83 kJ/mol    (from reference tables)

ΔHºf  (CO₂) = -393.52 kJ/mol  (from reference tables)

ΔHºf  (CO) = -110.53 kJ/mol (from referenfence tables)

Remember that ΔHreaction = ∑ ΔHºf products - ∑ΔHºf reactants

Part (a)

(1) C₂₁H₄₄ (s) + 32 O₂ (g) →  21 CO₂ (g) + 22 H₂O (g)

ΔHcomb₁ = {21(-393.52) +22(-241.83)} - { -476 } = -1.31 x 10⁴ kJ/mol

(2) C₂₁H₄₄ (s) + 43/2 O₂ (g) → 21 CO(g) + 22 H₂O(g)

ΔHcomb₂ = {21(-110.53) + 22(-241.83)} - {-476} = -7.17 x 10³ kJ/mol

(3) C₂₁H₄₄ (s) + 11 O₂(g)  →  21 C(graphite) + 22 H₂O(g)

ΔHcomb₃ = {22(-241.83)] - {-476} =  -4.84 x 10³ kJ/mol

Part (b)

We now have the heat of combustion for the reaction per mol of C₂₁H₄₄, so it is a simple matter to calculate the heat released for a 413 g candle once we convert it in number of moles.

MW C₂₁H₄₄ = 296.6 kJ/mol

n candle = m/MW = 413 g/296.6 g/mol = 1.39 mol

ΔH = ΔHcomb₁ x n = -1.31 x 10⁴ kJ/mol x 1.39 mol = -1.82 x 10⁴ kJ

Part (c)  

For the calculations for this part will use the three reactions for which we know their heats of combustion according to the data given:

5.19 % mass incompletely

6.50 % formation soot

100 - (5.19+6.50) = 88.31 %  complete reaction

So for  a 413 g candle we have the following number of moles reacting:

n₁ = 0.883 x 413 g/ 296.6 g/mol = 1.23 mol react completely

n₂ = 0.0519 x 413 g / 296.6 g/mol = 0.07 mol incomplete

n₃ = 0.065 x 413 g / 296.6 g/mol = 0.09 mol  soot formation

Now we can calculate the heat for the reactions:

ΔH₁ = ΔHcomb₁ x n₁ =  -1.31 x 10⁴ kJ/mol x 1.23 mol = -1.61 x 10⁴ kJ

ΔH₂ = ΔHcomb₂ x n₂ = -7.17 x 10³ kJ/mol x 0.07 mol = -5.02 x 10² kJ

ΔH₃ = ΔHcomb₃ x n₃ = -4.84 x 10³ kJ/mol x 0.09 mol = -4.36 x 10² kJ

q = ΔH₁ +ΔH₂ +ΔH₃  = -1.61 x 10⁴ kJ  + (-5.02 x 10² kJ) + (-4.36 x 10² kJ)

q = -1.70 x 10⁴ kJ

i believe it is c.) 1575 j.