Question

Оксид щелочного металла массой 56.4 г обработали раствором серной кислоты.Затем к нему прилили Нитрат бария, выпал осадок массой 46.6г. Какой это металл?1)Li2)Cs3)K4)Na

Answer 1

Это цезий.

R2O + H2SO4 = R2SO4 + H2O

R2SO4 + Ba(NO3)2 = BaSO4 + 2RNO3

- Рассчитаем количество вещества выпавшего осадка:
n(BaSO4) = m/M = 46,6г/233г/моль = 0,2моль

- По уравнению реакции:

n(R2SO4) = n(BaSO = 0,2моль

n(R2O) = n(R2SO4) = 0,2моль

- Рассчитаем молярную массу оксида:

M = m/n = 56,4г/0,2г/моль = 282 г/моль

Теперь составим уравнение:

2x + 16 = 282

2х = 266

х = 133 - Цезий

 

Answer 2

Identification of the Metal

To determine the metal in question, we need to analyze the chemical reactions that occurred during the experiment.

The given information states that an oxide of an alkali metal with a mass of 56.4 g was treated with sulfuric acid. Afterward, barium nitrate was added, resulting in the formation of a precipitate weighing 46.6 g.

To identify the metal, we need to consider the reactions involved and the solubility rules of alkali metal compounds.

Reaction Analysis

The reaction between the oxide of the alkali metal and sulfuric acid can be represented as follows:

Metal Oxide + Sulfuric Acid → Metal Sulfate + Water

Since the resulting compound is a sulfate, it is expected to be soluble in water.

The subsequent reaction involves the addition of barium nitrate to the solution:

Metal Sulfate + Barium Nitrate → Barium Sulfate + Metal Nitrate

Barium sulfate is insoluble in water and forms a precipitate. The mass of the precipitate is given as 46.6 g.

Solubility Rules

Based on the solubility rules, alkali metal sulfates are soluble in water, while barium sulfate is insoluble. This means that the precipitate formed is barium sulfate.

Conclusion

From the given information, we can conclude that the metal in question is barium (Ba).

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