Question

Какая масса гидроксида натрия образуется при взаимодействии с 2.3 граммами с водой

Answer 1

Получается:

1. n(Na) =2,3|23=0,1 (моль)
2. 2Na +2H2O =2NaOH + H2
2моль------------2 моль
0,1 моль---------х х=0,1 моль NaOH
(NaOH) =23 +16 +1 =40
m(NaOH) =0,1*40 =4 (г)

Answer 2

Calculating the Mass of Sodium Hydroxide Formed

When sodium hydroxide reacts with water, it forms sodium hydroxide solution. To calculate the mass of sodium hydroxide formed when 2.3 grams of sodium reacts with water, we need to consider the stoichiometry of the reaction.

The balanced chemical equation for the reaction of sodium with water is:

2 Na + 2 H2O -> 2 NaOH + H2

From the equation, we can see that 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide.

Given that the molar mass of sodium is 22.99 g/mol and the molar mass of water is 18.015 g/mol, we can calculate the amount of sodium hydroxide formed.

Using the molar masses: - Molar mass of NaOH = 22.99 (Na) + 15.9994 (O) + 1.00794 (H) = 39.99734 g/mol

Now, let's calculate the mass of sodium hydroxide formed: - Moles of Na = 2.3 g / 22.99 g/mol = 0.1 moles - Moles of NaOH formed = 0.1 moles

So, the mass of sodium hydroxide formed when 2.3 grams of sodium reacts with water is 3.999734 grams.

This calculation is based on the stoichiometry of the reaction and the molar masses of the substances involved.