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Stoichiometry Propane (C3H8) can be burned to produce heat for homes. The products of...

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annaeleanor | Student | eNoter

Posted May 10, 2013 at 8:48 PM via web

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Stoichiometry

Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to occur, the ratio of the amounts of propane to oxygen must be adjusted correctly.

a. Use the balanced equation C3H8 + 5O2 = 3CO2 + 4H2O to answer the following questions.

      i. How many moles of CO2 are produced from 5 moles O2?

      ii. How many grams of CO2 are produced from 5 moles O2?

      iii. How many grams of CO2 are produced from 128.00 g O2?

      iv. A total of 132.33 g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed. 

      • Determine if one of the reactants is a limiting reagent.

      • How many grams of CO2 and H2O will be produced?

      • How many grams of the nonlimiting reagent will be left unreacted?

2 Answers | Add Yours

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Kay Morse | College Teacher | (Level 1) Senior Educator

Posted May 10, 2013 at 11:02 PM (Answer #2)

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Complete combustion occurs when the oxygen level is plentiful. Lack of plentiful oxygen results in incomplete combustion. For complete propane oxygen combustion, the amount of oxygen is

5O2 to C3H8

This results in 

→ 3CO2 + 4H2O

propane + oxygen → carbon dioxide + water

C3H8 + 5O2 → 3CO2 + 4H2O

 

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Top Answer

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llltkl | College Teacher | Valedictorian

Posted May 11, 2013 at 3:38 AM (Answer #3)

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The balanced chemical reaction for the combustion is:


C3H8 + 5O2 → 3CO2 + 4H2O

From the reaction stoichiometry the answers follow:  i. 3

      ii. 3*44 = 132 grams.

      iii. 5*32 = 160 gram O2 produces 132 gram CO2

so, 128 gram O2 should produce 132*128/160 = 105.6 gram CO2.

    iv. (3*12.0+8*1.0) = 44 gram propane reacts with 5*32 = 160 gram oxygen.

So. 132.33 gram propane should react with 160*132.33/44 gram oxygen = 481.2 gram O2. But actually 384.0 gram O2 is present. So, O2 must be the limiting reagent here.

Grams of CO2 formed in this case = 132*384/160 = 316.8.

Grams of H2O formed in this case = 4*(16+2)*384/160 = 172.8

Amount of propane reacted = 44*384/160 = 105.6 gram

Amount of propane (non limiting reagent) remaining unreacted = 132.33-105.6 = 26.73 grams.

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