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pifojohi123
@pifojohi123
August 2022
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Сколько граммов муравьинно-этилового эфира выделится при взаимодействии 90г 20% муравьиной кислоты и 40г 40% этилового спирта?
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mrvladimir2
Verified answer
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Alexei78
Дано
m(ppa CHOOH)=90 g
W(CHOOH)=20%
m(ppaC2H5OH)=40 g
W(C2H5OH)=40%
----------------------
m(CHOOC2H5)-?
m(CHOOH)= 90*20%/100% = 18 g
M(CHOOH)=46 g/mol
n(CHOOH) = m/M=18/46=0.39 mol
m(C2H5OH)=40*40%/100% = 16 g
M(C2H5OH)=46 g/mol
n(C2H5OH)=m/M=16/46=0.35 mol
n(CHOOH)>n(C2H5OH)
CHOOH+C2H5OH-->CHOOC2H5+H2O M(CHOOC2H5)=74 g/mol
n(C2H5OH)=n(CHOOC2H5)= 0.35 mol
m(CHOOC2H5)= n*M = 0.35*74=25.9 g
ответ 25.9 г
1 votes
Thanks 1
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Verified answer
.....................................................................m(ppa CHOOH)=90 g
W(CHOOH)=20%
m(ppaC2H5OH)=40 g
W(C2H5OH)=40%
----------------------
m(CHOOC2H5)-?
m(CHOOH)= 90*20%/100% = 18 g
M(CHOOH)=46 g/mol
n(CHOOH) = m/M=18/46=0.39 mol
m(C2H5OH)=40*40%/100% = 16 g
M(C2H5OH)=46 g/mol
n(C2H5OH)=m/M=16/46=0.35 mol
n(CHOOH)>n(C2H5OH)
CHOOH+C2H5OH-->CHOOC2H5+H2O M(CHOOC2H5)=74 g/mol
n(C2H5OH)=n(CHOOC2H5)= 0.35 mol
m(CHOOC2H5)= n*M = 0.35*74=25.9 g
ответ 25.9 г