Question

К раствору, содержащиму 20 грам гидроксида натрия, прилили раствор, содержащий 33 грамма хлороводорода. Какая масса соли образовалась? помогите пожалуйста!!!

Answer 1

пишем уровнение реакции. вычисляем моли 
ню гидроксида натрия равна 0.5моль
ню хлороводорода равно  0.8 моль

так как гн взто в избытке считаем по хлороводороду

х=20*58.5/40=29.25 г
 

Answer 2

Calculating the Mass of Salt Formed

To calculate the mass of salt formed when a solution containing 20 grams of sodium hydroxide is mixed with a solution containing 33 grams of hydrochloric acid, we can use the concept of a chemical reaction between the two compounds. The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) results in the formation of water (H2O) and sodium chloride (NaCl).

The balanced chemical equation for this reaction is: NaOH + HCl → NaCl + H2O

According to the stoichiometry of the balanced chemical equation, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl. From the molar masses of the compounds involved, we can calculate the mass of salt formed.

Calculation Steps

1. Calculate the molar mass of NaOH and HCl. 2. Determine the limiting reactant. 3. Use the amount of the limiting reactant to calculate the amount of salt formed.

Molar Mass Calculation

The molar mass of NaOH (sodium hydroxide) is approximately 40 g/mol, and the molar mass of HCl (hydrochloric acid) is approximately 36.5 g/mol.

Limiting Reactant

To determine the limiting reactant, we compare the amount of each reactant to the stoichiometry of the balanced chemical equation. In this case, the limiting reactant is the one that is completely consumed in the reaction.

Calculation

Given: - Mass of NaOH = 20 grams - Mass of HCl = 33 grams

First, we need to convert the masses of NaOH and HCl to moles using their respective molar masses: - Moles of NaOH = 20 g / 40 g/mol = 0.5 moles - Moles of HCl = 33 g / 36.5 g/mol ≈ 0.904 moles

Since NaOH and HCl react in a 1:1 ratio, HCl is the limiting reactant because it is completely consumed when 0.5 moles of NaOH react.

The amount of salt formed can be calculated using the moles of the limiting reactant: - Moles of NaCl formed = 0.5 moles

Finally, we convert the moles of NaCl to grams using the molar mass of NaCl (approximately 58.5 g/mol): - Mass of NaCl formed = 0.5 moles * 58.5 g/mol ≈ 29.25 grams

The mass of salt formed is approximately 29.25 grams.