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IvanRuds
@IvanRuds
April 2021
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Помогите пожалуйста.Найти массу раствора naoh 15% если при его взаимодействии с alcl3 образовалось 60 грамм осадка
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wasurenagusa
AlCl3 + 3NaOH = Al(OH)3 + 3NaCl
n(Al(OH)3) = 60/78 = 0,77 моль
n(NaOH) = 3*
0,77 = 2,31 моль
m(
NaOH) =
2,31*40 = 92,4 г
m р-ра = m в-ва*100%/
ω% =
92,4*100%/15% = 616 г
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Answers & Comments
n(Al(OH)3) = 60/78 = 0,77 моль
n(NaOH) = 3*0,77 = 2,31 моль
m(NaOH) = 2,31*40 = 92,4 г
m р-ра = m в-ва*100%/ω% = 92,4*100%/15% = 616 г