$$ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} $$

In the reaction represented above, what is the total number of moles of reactants consumed when 1.00 mole of \(\ce{CO2(g)}\) is produced?

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We need 1 mole of \(\ce{C3H8}\) and 5 moles of \(\ce{O2}\) to form 3 moles of \(\ce{CO2}\).

To maintain this ration, we would need \(\dfrac{1}{3}\) mole of \(\ce{C3H8}\) and \(\dfrac{5}{3}\) moles of \(\ce{O2}\) to form 1 mole of \(\ce{CO2}\)

$$ \frac{1}{3} + \frac{5}{3} = \frac{6}{3} = \pu{2 mol reactants} $$