Question

36г води взаємодіють з Натрій оксидом. Обчисліть масу утвореної сполуки.

Answer 1

Na2O + H2O = 2NaOH
m(H2O) = 36 г

m(NaOH) - ?

Решение:

1) μ(H2O) = 36 г / 18 г/моль = 2 моль

2) μ(NaOH) = 2 моль * 2 = 4 моль

3) m(NaOH) = 4 моль * 40 г/моль = 160 г

Ответ: 160 г

Answer 2

Interaction of Water with Sodium Oxide

When water interacts with sodium oxide (Na2O), it forms sodium hydroxide (NaOH) according to the following chemical equation:

Na2O + H2O → 2NaOH

This reaction produces sodium hydroxide as the compound.

Calculating the Mass of the Compound

To calculate the mass of the compound formed, we can use the concept of stoichiometry. The molar mass of sodium oxide (Na2O) is approximately 61.9789 g/mol, and the molar mass of water (H2O) is approximately 18.01528 g/mol.

From the balanced chemical equation, we can see that 1 mole of sodium oxide reacts with 1 mole of water to produce 2 moles of sodium hydroxide. Therefore, the molar mass of sodium hydroxide (NaOH) is approximately 39.9971 g/mol.

Given the molar mass of sodium oxide and water, we can calculate the mass of sodium hydroxide formed when 36g of water interacts with sodium oxide.

Using the molar masses and the stoichiometry of the reaction, the mass of sodium hydroxide formed can be calculated as follows:

36g H2O × (1 mol H2O / 18.01528 g) × (2 mol NaOH / 1 mol H2O) × (39.9971 g / 1 mol NaOH) ≈ X g NaOH

By performing the calculations, we can determine the mass of sodium hydroxide formed when 36g of water interacts with sodium oxide.

This calculation will give us the mass of the compound formed as a result of the interaction between water and sodium oxide.

Let's calculate the value of X.