Answer :
Moles of o2 gas would be consumed in the complete combustion of 0.100 mol of c5h12 is 0.800mol of O2.
What is combustion?
Combustion, or burning, is high-temperature exothermic redox chemical reaction between a fuel (the reductant) and an oxidant, usually atmospheric oxygen, that produces oxidized, often gaseous products, in a mixture termed as smoke. Combustion does not always result in fire, because a flame is only visible when the substances undergoing combustion vaporize, but when it does, a flame is a characteristic indicator of the reaction. While activation energy must be overcome to initiate the combustion (e.g., using a lit match to light fire), the heat from a flame may provide enough energy to make reaction self-sustaining.
Combustion is often complicated sequence of elementary radical reactions. Solid fuels, such as wood and coal, first undergo endothermic pyrolysis to produce the gaseous fuels whose combustion then supplies the heat required to produce more of them. Combustion is often hot enough that incandescent light in form of either glowing or a flame is produced. A simple example can be seen in combustion of hydrogen and oxygen into water vapor, a reaction which is commonly used to fuel rocket engines.
We're asked to find number of moles of O2 needed to completely combust 1mol C5H12.
To do this, we first need to write chemical equation for this reaction, following the general formula for a hydrocarbon combustion reaction:
hydrocarbon +oxygen →carbon dioxide +water (vapor)
So,
C5H12(l)+O2(g)→CO2(g)+H2O(g)(unbalanced)
Let's now balance the equation. We'll first put a "5" in front of the CO2
to balance carbons:
C5H12(l)+O2(g)→5CO2(g)+H2O(g)(unbalanced)
Now we'll put a "6" in the front of H2O
to balance hydrogens:
C5H12(l)+O2(g)→5CO2(g)+6H2O(g)(unbalanced)
We now have 2 oxygens on left side and 16 on the right side. Putting an "8" in the front of O2
balances the chemical equation:
C5H12(l)+8O2(g)→5CO2(g)+6H2O(g)
Now, we can realize that stoichiometric ratio of pentane to oxygen is 1:8 (from coefficients).
Therefore, in order to completely combust 1mol of pentane, we must use
8mol of O2
so, for 0.100 mol C5H12 require 0.100*8mol O2 i.e. 0.800mol O2.
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