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Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
  • CO₂: 3 moles* 44 g/mole= 132 grams
  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

[tex]mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}[/tex]

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

[tex]mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}[/tex]

mass of H₂O= 183.6 grams

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

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