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A mixture of methanol and methyl acetate contains 7.0 weight percent methanol. Determine the number of gmols of methanol in 300.0 kilograms of the mixture.

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daigz92

Answer:

656,25 moles of CH3OH

Explanation:

If the mixture has 7% of weight in methanol (CH3OH), it means that for every 100 kg of the mixture there are 7 kg of methanol.

To solve the problem we just use this relation and convert the kilograms of methanol to gmoles of methanol using the molecuar weight of the methanol (32g/mol):

[tex]300 kgmixture*\frac{7kgCH3OH}{100kg}*\frac{1000g}{1kg}*\frac{1molCH3OH}{32g}=656,25molesCH3OH[/tex]

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