Відповідь:
w(примесей)=2.1%
Пояснення:
Дано: m(тех.C₆H₅NH₂)= 19 г , m(C₆H₂Br₃-NH₂)=66 г . w(примесей)-?
I. Запишем реакцию:
C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃-NH₂ +3HBr
С реакции видно, что соотношение количества между C₆H₂Br₃-NH₂ и C₆H₅NH₂, как 1:1. По этому, n(C₆H₅NH₂)=n(C₆H₂Br₃-NH₂)
II.Проведем вычисления:
w(E)=m(E)/m(раствора) *100%
n(E)=m(E)/Mr(E)
Сделаем соотношение:
m(C₆H₅NH₂)/Mr(C₆H₅NH₂)=m(C₆H₂Br₃-NH₂)/Mr(C₆H₂Br₃-NH₂)
X / 93 = 66/330
X = 66*93 / 330 = 18.6
m(примесей)=19-18.6=0.4 г
w(примесей)=0.4/19 *100%=2.1%
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Поделитесь своими знаниями, ответьте на вопрос:
При растворении n2co3 в избытке соляной кислоты выделялось 4, 48 л. какого то газа. чему равна масса растворимого n2co3
что за "n" азот? такое бывает?
x 4,48
n2co3 + 2hcl = ncl2 + h2o + co2(газ)
88 22,4
х= (4,48*88) \ 22,4 = 17,6 г.