Предмет: Химия, автор: hgfdrafa

Какова масса соли, получившийся при взаимодействии 100г 10% раствора NaOH с хлоридом аммония

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Автор ответа: анонимсм

0

NaOH+NH4Cl=NaCl+NH3+H2O
m(NaOh)=100*0.1=10 г
тогда n(NaOH)=0.25 моль
а маса соли=0.25*58.5=14,625 г

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