Question

What mass, in grams, of CO2 and H2O
is formed from 2.55 mol of propane?

Answer 1

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

Step-by-step 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?

`mass of CO_(2) =(2.55 moles of C_(3) H_(8)*132 gramsof CO_(2) )/( 1 mole of C_(3) H_(8))`

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?

`mass of H_(2)O =(2.55 moles of C_(3) H_(8)*72 gramsof H_(2)O )/( 1 mole of C_(3) H_(8))`

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.