Question

Какая масса соли и воды образуется при взаимодействии 15г NaOH и 13г HCL

Answer 1

Ответ:

NaOH + HCl = NaCl + H2O

n (NaOH) = 15г / 40 г/моль = 0,375

n (HCl) = 13г / 36,5 г/моль = 0,356

Соляная кислота находится в небольшом недостатке, считаем по ней.

n (NaCl) = n (HCl) = 0,356 моль

n (NaCl) = 0,356 моль × 58,5 г/моль = 20,8 грамм

Answer 2

Reaction and Product Formation

When 15g of NaOH and 13g of HCl react, the following reaction occurs:

NaOH + HCl → NaCl + H2O

The balanced chemical equation for this reaction is:

NaOH + HCl → NaCl + H2O.

Calculation of Product Masses

To calculate the masses of the products formed, we can use the concept of stoichiometry. The molar mass of NaOH is approximately 40 g/mol, and the molar mass of HCl is approximately 36.5 g/mol

First, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product formed.

The molar amount of NaOH is calculated as follows: 15g NaOH ÷ 40 g/mol = 0.375 mol NaOH

The molar amount of HCl is calculated as follows: 13g HCl ÷ 36.5 g/mol = 0.356 mol HCl

Since the molar amount of HCl is less than that of NaOH, HCl is the limiting reactant.

Using the stoichiometry of the reaction, we can calculate the amount of NaCl and H2O formed. The molar ratio of NaOH to NaCl is 1:1, and the molar ratio of HCl to H2O is 1:1.

Therefore, the amount of NaCl formed is 0.356 mol, and the amount of H2O formed is also 0.356 mol.

Finally, we can calculate the masses of NaCl and H2O formed:

Mass of NaCl = 0.356 mol × 58.5 g/mol = 20.76g Mass of H2O = 0.356 mol × 18 g/mol = 6.408g

So, when 15g of NaOH and 13g of HCl react, approximately 20.76g of NaCl and 6.408g of H2O are formed.

This calculation assumes ideal conditions and complete reaction conversion.