The combustion of methanol (ch4o) is described by this balanced chemical equation: 2ch4o(g) + 3o2(g) → 4h2o(g) + 2co2(g) methanol and oxygen (o2) will always react in a ratio of

Methanol and oxygen will always react in a ratio of 2:3

Explanation:

According to the law of conservation of mass, the mass of products must  be same as the mass of reactants and thus the number of atoms on the product side must be equal to the number of atoms on the reactant side.

Thus methanol and oxygen will always react in the mole ratio of 2:3.

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