In the combustion reaction of propane, C3H8 + 5O2 -> 3 CO2 + 4 H2O, if starting

Question

In the combustion reaction of propane, C3H8 + 5O2 -> 3 CO2 + 4 H2O, if starting with 1.2 miles of propane and 6 moles of oxygen, how many moles of water can be made? In the combustion reaction of propane, C3H8 + 5O2 -> 3 CO2 + 4 H2O, if starting with 1.2 miles of propane and 6 moles of oxygen, how many moles of water can be made? In the combustion reaction of propane, C3H8 + 5O2 -> 3 CO2 + 4 H2O, if starting with 1.2 miles of propane and 6 moles of oxygen, how many moles of water can be made?

Explanation / Answer

the given reaction is

C3H8 + 502 ----> 3C02 + 4H20

now

given

1.2 moles of C3H8 , 6 moles of 02

consider the reaction

C3H8 + 502 ---> 3C02 + 4H20

from the above reaction

moles of 02 required = 5 x moles of C3H8

so

moles of O2 required = 5 x 1.2

moles of O2 required = 6

so 6 moles of O2 is required

so none of the reactants is in excess

now

C3H8 + 502 ---> 3C02 + 4H20

we can see that

moles of H20 formed = 4 x moles of C3H8 reacted

so

moles of H20 formed = 4 x 1.2

moles of H20 formed = 4.8

so

4.8 moles of water can be made

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