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Question

Consider the following combustion reaction:
C3H8(g)+5O2(g)3CO2(g)+4H2O(g)
At STP (atm), when 64.4 L of propane is reacted with 87.0 L of oxygen gas, volume of the CO2 gas produced is:

A
105 L of CO2
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B
258 L of CO2
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C
39.2 L of CO2
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D
52.2 L of CO2
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Solution

The correct option is D 52.2 L of CO2
1 mole of propane requires 5 moles of oxygen
22.4 L of propane requires (5×22.4) L of oxygen
64.4 L of propane will require 5×22.422.4×64.4=322 L of oxygen
Therefore, oxygen is the limiting reagent.
5 moles of oxygen produce 3 moles of CO2
(5×22.4) L of oxygen produces (3×22.4) L of CO2
So, 87 L of oxygen will produce 3×22.45×22.4×87= 52.2 L of CO2

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