How many molecules of co2. h2o. c3h8. and o2 will be present if the reaction goes to completion?

consider the mixture of propane, C3H8, and O2 (figure shows 10 O2 molecules and 3 propane molecules).

C3H8 + 5O2 ---> 3CO2 + 4H2O

O2 - limiting reactant

how many molecules of CO2, H2O, C3H8, and O2 will be present if the reaction goes to completion?

3,4,0,0 - incorrect; 6,8,5,0 - incorrect; 6,8,0,0 - incorrect

thank you

Answer

                     C3H8 +    5O2         --->    3CO2     +     4H2O

Initial moles     3               10                      0                    0

Limiting reagent = oxygen

As per the stoichiometry of reaction

1 mole of propane will react with 5 moles of oxygen

so 1 mole of oxygen will react with 1/5 moles of propane

so 10 moles of oxygen will react with 10/ 5 moles of propane = 2 moles of propane

Again

As per the stoichiometry, 5 moles of oxygen will react with 1 mole of propane to give 3 moles of CO2 and 4 moles of H2O

so 10 moles of oxygen will react with 2 moles of propane ( so one mole of propane will remain) to give 6 moles of CO2 and 8 moles of H2O

So the correct answer is

CO2 = 6

H2O = 8

Propane = 1

oxygen = 0

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