teresaduggan7116 teresaduggan7116

10-11-2017
Chemistry

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Given 8.05 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

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Kalahira Kalahira

21-11-2017

Answer: CH3CH2CH2COOH + CH3CH2OH = CH3CH2CH2COOCH2CH3 + H2O moles acid = 8.05 g /88.108 g/mol=0.0800 moles ester = 0.0800 mass ester = 0.0800 mol x 116.162 g/mol=9.29 g % yield = 5.10 x 100/ 9.29=54.9 moles ester = 0.0800 x 78/100=0.0624 mass ester = 0.0624 x 116.162 = 7.25 g

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