Question

4 г натрію прореагували з сульфатною кислотою. Знайти масу солі що утворилася, та об'єм водню.

Answer 1

В приложении решение...................................................

Answer 2

Reaction of 4g of Sodium with Sulfuric Acid

When 4 grams of sodium react with sulfuric acid, the following chemical reaction occurs:

2Na + H2SO4 -> Na2SO4 + 2H2

This reaction produces sodium sulfate (Na2SO4) and hydrogen gas (H2).

Mass of Sodium Sulfate Formed

The molar mass of sodium sulfate (Na2SO4) is approximately 142.04 g/mol. To find the mass of sodium sulfate formed, we can use the stoichiometry of the reaction.

From the balanced chemical equation, we can see that 2 moles of sodium react with 1 mole of sulfuric acid to produce 1 mole of sodium sulfate. This means that 2 moles of sodium (2 * 23 = 46 g) react with 1 mole of sulfuric acid to produce 142.04 g of sodium sulfate.

Using the concept of stoichiometry, we can calculate the mass of sodium sulfate formed when 4 grams of sodium react with sulfuric acid:

(4 g Na) * (1 mol Na / 23 g Na) * (1 mol Na2SO4 / 2 mol Na) * (142.04 g Na2SO4 / 1 mol Na2SO4) ≈ 31.0 g of sodium sulfate formed.

Volume of Hydrogen Gas Produced

The balanced chemical equation shows that 2 moles of hydrogen gas are produced for every 1 mole of sulfuric acid that reacts. Using the ideal gas law, we can calculate the volume of hydrogen gas produced.

Given: - 1 mole of any gas occupies approximately 22.4 liters at standard temperature and pressure (STP).

Using the stoichiometry of the reaction, we can calculate the moles of hydrogen gas produced:

(4 g Na) * (1 mol Na / 23 g Na) * (2 mol H2 / 2 mol Na) ≈ 0.174 moles of H2.

At STP, 1 mole of gas occupies 22.4 liters. Therefore, the volume of hydrogen gas produced is approximately 3.9 liters.

In summary, when 4 grams of sodium react with sulfuric acid, approximately 31.0 grams of sodium sulfate are formed, and the volume of hydrogen gas produced is approximately 3.9 liters.