Question

В реакции алюминия с серной кислотой образовалось 75,5 г сульфата алюминия. Вычислить массу и количество вещества алюминия, вступившего в реакцию.

Answer 1

Решение в приложении.....

Answer 2

Может немного не понятно, но решение учителей обрадует)

Answer 3

Reaction between Aluminum and Sulfuric Acid

When aluminum reacts with sulfuric acid (H2SO4), it forms aluminum sulfate (Al2(SO4)3) and hydrogen gas (H2) as products. The balanced chemical equation for this reaction is:

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2

To calculate the mass and amount of aluminum that reacted, we need to use the given information that 75.5 grams of aluminum sulfate (Al2(SO4)3) was formed.

Calculating the Mass of Aluminum

To calculate the mass of aluminum, we can use the stoichiometry of the balanced chemical equation. According to the equation, 2 moles of aluminum react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. From the molar mass of aluminum sulfate (342.15 g/mol), we can calculate the moles of aluminum sulfate formed:

75.5 g Al2(SO4)3 * (1 mol Al2(SO4)3 / 342.15 g Al2(SO4)3) = 0.2207 mol Al2(SO4)3

Since the stoichiometry of the balanced equation is 2 moles of aluminum to 1 mole of aluminum sulfate, we can calculate the moles of aluminum:

0.2207 mol Al2(SO4)3 * (2 mol Al / 1 mol Al2(SO4)3) = 0.4414 mol Al

To calculate the mass of aluminum, we can use the molar mass of aluminum (26.98 g/mol):

0.4414 mol Al * 26.98 g Al/mol = 11.90 g Al

Therefore, the mass of aluminum that reacted is 11.90 grams.

Calculating the Amount of Aluminum

The amount of aluminum can be calculated using the Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 particles. In this case, we are calculating the moles of aluminum:

0.4414 mol Al * (6.022 x 10^23 particles/mol) = 2.657 x 10^23 particles of aluminum

Therefore, the amount of aluminum that reacted is 2.657 x 10^23 particles.

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