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Ответ:

Medvedevalilit2004

08.07.2022 06:19

Відповідь:

1. 50.5 % | 2. 110 g .

Пояснення:

1. Дано:

m(CH₃COOH)= 90 g | CH₃COOH + HOCH₃ ---> CH₃COOCH₃ + H₂O

m(CH₃OH)=64 g | n(CH₃COOH)= 90 / 60 = 1.5 mol

mпр(CH₃COOCH₃)=100g| n(CH₃OH)= 64 / 32 = 2 mol

| CH₃OH - многа на 0.5 моль - расчет за уксусом

w(CH₃COOCH₃) - ? | n(CH₃COOCH₃)=n(CH₃COOH)=1.5 mol

| m(CH₃COOCH₃)=132*1.5= 198 g

w(вихода) (CH₃COOCH₃) = m практическая / m теоритическая * 100%

w(вихода) (CH₃COOCH₃)= 100/192 *100 % = 50.5 %

2. Дано:

n(CO₂)= 5 mol | CH₃COOC₂H₅ + 5O₂ --> 4CO₂ + 4H₂O

| с 1 моль ефира - 4 моль угл. газа

m(CH₃COOC₂H₅) -? | с х моль ефира - 5 моль угл. газа

1 / 4 = x / 5 ---> x=5/4=1.25 mol

m=n*Mr. m(CH₃COOC₂H₅)= 1.25 * 88 = 110 g

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