Вычисли, какая масса алюминия потребуется для реакции с 0,12 моль кислорода. (Введи ответ с точностью до двух знаков после запятой). Ответ: m(Al)=

Вычисли, какая масса алюминия потребуется для реакции с 0,12 моль кислорода.

(Введи ответ с точностью до двух знаков после запятой).

Ответ: m(Al)=
- Предмет:
  Химия
- Автор:
  azariaandersen
- 5 лет назад

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Спасибо
- Автор:
  judy
- 5 лет назад
- 0
Дано:n(O₂) = 0,12 моль+AlНайти:m(Al)Решение:4Al + 3O₂ = 2Al₂O₃По уравнению реакции алюминия 4 моль, кислорода 3 моль.По условию алюминия х моль, кислорода 0,12 моль.Составим уравнение: n(Al) = $\frac{0,12*4}{3} =$ 0,16 мольMr(Al) = 27 г/мольm(Al) = n(Al)*Mr(Al) = 0,16*27 = 4,32гОтвет: m(Al) = 4,32г
- Автор:
  juliánd0ys
- 5 лет назад
- 0
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Вычисли, какая масса алюминия потребуется для реакции с 0,12 моль кислорода. (Введи ответ с точностью до двух знаков после запятой). Ответ: m(Al)=

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Вычисли, какая масса алюминия потребуется для реакции с 0,12 моль кислорода.

(Введи ответ с точностью до двух знаков после запятой).

Ответ: m(Al)=