In the complete combustion of hexane (C6H14), how many moles of CO2 will be made when 15.7 moles of hexane are burned?
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In every complete combustion equation a hydrocarbon `(C_xH_y)` reacts with oxygen `(O_2)` to produce water `(H_2O)` and carbon dioxide `(CO_2)` . Combustion reactions shows the pattern of :
`C_xH_y + O_2(g) rarr H_2O(g) + CO_2(g) `
So, the balanced chemical equation of the complete combustion reaction of hexane `(C_6H_14)` is:
`C_6H_14 + 19/2 O_2 rarr 6CO_2 + 7H_2O`
From the stoichiometry of the above reaction it is evident that,
1 mole of hexane on complete combustion produces 6 moles of `CO_2`
Hence, 15.7 moles of hexane on complete combustion produces 6*15.7 = 94.2 moles of `CO_2`
Therefore, 94.2 moles of `CO_2` will be made when 15.7 moles of hexane are burned.
The balanced reaction for the combustion of hexane is:
`C_6H_(14) + 19/2O_2 -> 7H_2O + 6CO_2`
Use stoichiometry to find the amount of CO2 produced with 15.7 moles of C6H14.
`15.7"mol"C_6H_(14) = (6"mol"CO_2)/(1"mol"C_6H_(14)) = 94.2"mol"CO_2`
Remember, there should be three sig figs!
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