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drek231 [11]
3 years ago
7

Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su

pply of air
Chemistry
1 answer:
Ratling [72]3 years ago
5 0

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

<em>1</em> C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • C₅H₁₂ is the only <em>reactant</em> that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
  • CO₂ is the only <em>product</em> that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

<em>1 </em>C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + ? H₂O

  • Similarly, C₅H₁₂ is the only <em>reactant</em> that contains H atoms. H₂O is the only <em>product</em> that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

<em>1</em> C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

  • Both CO₂ and H₂O are <em>products</em> that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the <em>product</em> side corresponds to 8 O₂ molecules on the reactant side.

<em>1</em> C₅H₁₂ + <em>8</em> O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

  • All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", <em>GCSE Chemistry (Single Science)</em>, BBC Bitesize.

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Answer:

a)

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0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

(F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = 23.562

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the higher heating values (HHV)_f per unit mass of LPG = 49.9876 MJ/kg

the lower heating values (LHV)_f per unit mass of LPG = 46.4933 MJ/kg

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The stoichiometric equation can be expressed as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2  \ + \ bH_2O \ + \ cN_2

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(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

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(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

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2x = 2a + b

x = \frac{2a+b}{2} \\ \\ x =  \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95

Equating the coefficient of Nitrogen

c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612

Therefore, The overall  balanced combustion  reaction can now be written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

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(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\  (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\  (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562

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(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg

calculating the lower heating values (LHV)_f per unit mass of LPG; we have:

(LHV)_f = (HHV)_f - \delta H_w \\ \\  (LHV)_f = (HHV)_f  - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f   = 49.9876 \ MJ/kg -  [\frac{3.8*18}{44.2}*2.258 \ MJ/kg]  \\ \\ (LHV)_f = 46.4933 \ M/kg

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