Предмет: Химия, автор: Julia6878

Найти массу Al4C3,если 12 г алюминия
взаимодействует с 18 г углерода

Ответы

Автор ответа: saliva
0
  12г     18г    хг
4Al + 3C = Al4C3
1) Находим количество в-ва алюминия
n=m/M = 12/ 27 = 0,44 моль
2) Находим количество в-ва углерода
n=m/M= 18/12 = 1,5 моль 
.Расчет ведем по алюминию
n(карбида)= 0,25*n(Al) = 0,11 моль
m(карбида)=n*M = 0,11 * 144 = 15,84 г 
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