Предмет: Химия, автор: 3456725

найти объем водорода, если в реакцию с алюминием вступало 73 г 10% раствора соляной кислоты

Ответы

Автор ответа: Potassium

10% от 73-х = 7.3.

2Al + 6HCl = 2AlCl3 + 3H2

Над соляной кислотой ставим 7.3, внизу мол.массу * 6 - 36.5 * 6 = 219.
Над водородом Х, внизу молярный объём, умноженный на 3 - 22.4 * 3 = 67.2.

Имеем пропорцию:

7.3/219 = X/67.2

X = (7.3*67.2)/219 = 2.24

Ответ: выделилось 2.24 литра водорода.

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