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Limiting reagent
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=== Example for two reactants === Consider the [[combustion]] of [[benzene]], represented by the following [[chemical equation]]: :<chem>2 C6H6(l) + 15 O2(g) -> 12 CO2(g) + 6 H2O(l)</chem> This means that 15 [[Mole (unit)|moles]] of molecular [[oxygen]] (O<sub>2</sub>) is required to react with 2 moles of benzene (C<sub>6</sub>H<sub>6</sub>) The amount of oxygen required for other quantities of benzene can be calculated using [[cross-multiplication]] (the rule of three). For example, if 1.5 mol C<sub>6</sub>H<sub>6</sub> is present, 11.25 mol O<sub>2</sub> is required: :<math chem> 1.5 \ \ce{mol\,C6H6} \times \frac{15 \ \ce{mol\,O2}}{2 \ \ce{mol\,C6H6}} = 11.25 \ \ce{mol\,O2}</math> If in fact 18 mol O<sub>2</sub> are present, there will be an excess of (18 - 11.25) = 6.75 mol of unreacted oxygen when all the benzene is consumed. Benzene is then the limiting reagent. This conclusion can be verified by comparing the mole ratio of O<sub>2</sub> and C<sub>6</sub>H<sub>6</sub> required by the balanced equation with the mole ratio actually present: * required: <math chem>\frac{\ce{mol\,O2}}{\ce{mol\,C6H6}} = \frac{15 \ \ce{mol\,O2}}{2 \ \ce{mol\,C6H6}}=7.5 \ \ce{mol\,O2}</math> * actual: <math chem>\frac{\ce{mol\,O2}}{\ce{mol\,C6H6}} = \frac{18 \ \ce{mol\,O2}}{1.5 \ \ce{mol\,C6H6}}=12 \ \ce{mol\,O2}</math> Since the actual ratio is larger than required, O<sub>2</sub> is the reagent in excess, which confirms that benzene is the limiting reagent.
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