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choli [55]
4 years ago
5

Given 7.90 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

% yield?
Chemistry
1 answer:
Lana71 [14]4 years ago
7 0
Answer : 9.293 g

Explanation : The reaction is 

C_{4}H_{8}O_{2}  + C_{2}H_{5}OH ----\ \textgreater \  C_{6}H_{12}O_{2} + H_{2}O

Here, if we assume the stoichiometric esterification ratio as 1:1 then,

no. of moles of  C_{4}H_{8}O_{2} = 7.05 g/ 88 = 8.011 X 10^{-2} moles

so here, no. of moles of ethyl butyl rate = no. of moles of butyric acid ;

\frac{m (C_{4}H_{8}O_{2})}{M (C_{4}H_{8}O_{2})} =  \frac{m (C_{6}H_{12}O_{8})}{M (C_{6}H_{12}O_{8})}

m(C_{6}H_{12}O_{8})) = (7.05 /88.11) X 116.16 = 9.29 g 
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Let me know if you need anything else. :)

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