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levcenkonina66
@levcenkonina66
September 2021
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Помогите органическая химия
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2Artyom4
Дано:
m (CH3COOC2H5) = 44 г
m пр (CH3CONH2) = 25 г
w (CH3CONH2) - ?
NH3 + CH3COOC2H5 = CH3CONH2 + C2H5OH
M (CH3COOC2H5) = 88 г/моль
M (CH3CONH2) = 59 г/моль
1) n (CH3CONH2) = n (CH3COOC2H5) = 44/88 = 0,5 моль
2) m теор (CH3CONH2) = 0,5*59 = 29,5 г
3) w (CH3CONH2) = 25/29,5*100% = 84,75%
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Answers & Comments
m (CH3COOC2H5) = 44 г
m пр (CH3CONH2) = 25 г
w (CH3CONH2) - ?
NH3 + CH3COOC2H5 = CH3CONH2 + C2H5OH
M (CH3COOC2H5) = 88 г/моль
M (CH3CONH2) = 59 г/моль
1) n (CH3CONH2) = n (CH3COOC2H5) = 44/88 = 0,5 моль
2) m теор (CH3CONH2) = 0,5*59 = 29,5 г
3) w (CH3CONH2) = 25/29,5*100% = 84,75%