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21 January, 01:42

The equation for the complete combustion CH₄ is: CH₄ + 2O₂ → CO₂ + 2H₂O If 3.70 moles of methane reacted with 5.40 moles of oxygen, what is the limiting reactant in the formation of water?

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  1. 21 January, 03:54
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    CH₄ + 2O₂ → CO₂ + 2H₂O

    from equation 1 mol 2 mol

    given 3.70 mol 5.40 mol

    1 mol CH₄ needs 2 mol O₂

    3.70 mol CH₄ need x mol O₂

    x=3.70*2/1=7.40 mol O₂

    So, we see that 3.70 mol CH₄ need 7.40 mol O₂, but only 5.40 mol O₂ given (it is not enough).

    So, O₂ is the limiting reactant.
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