titov-es3095
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При растворении n2co3 в избытке соляной кислоты выделялось 4, 48 л. какого то газа. чему равна масса растворимого n2co3

Химия

Ответы

A2017

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.

   

lyubavalev8

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%

*Поставьте лучшее

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При растворении n2co3 в избытке соляной кислоты выделялось 4, 48 л. какого то газа. чему равна масса растворимого n2co3
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