Question

Вычислите массу осадка,образовавшегося при взаимодействии двух растворов, один из которых содержит 21,2 г карбоната натрия, а второй-хлорид кальция

Answer 1

Записать уравнение реакции, над веществами записываем то что изве
стно и то что нужно найти, под формулами веществ записываем молеку
лярные массы карбоната натрия и карбоната кальция (осадка)
  21.2г                           Х г
  Na2CO3 + CaCl2= CaCO3 + 2NaCl
      106г                       100г                           
по пропорции находим Х=21.2*100/106=20 г

Answer 2

Calculating the Mass of Precipitate Formed

To calculate the mass of the precipitate formed when two solutions react, we can use the stoichiometry of the reaction and the principles of chemical reactions.

The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and calcium chloride (CaCl2) is:

Na2CO3 + CaCl2 → 2NaCl + CaCO3

From the balanced equation, we can see that 1 mole of calcium chloride reacts with 1 mole of sodium carbonate to produce 1 mole of calcium carbonate (the precipitate) and 2 moles of sodium chloride.

Given: - Mass of sodium carbonate (Na2CO3) = 21.2 g - Calcium chloride (CaCl2)

We need to calculate the mass of the precipitate (calcium carbonate, CaCO3) formed.

Solution

1. Calculate the molar mass of sodium carbonate (Na2CO3): - Molar mass of Na = 22.99 g/mol - Molar mass of C = 12.01 g/mol - Molar mass of O = 16.00 g/mol - Molar mass of Na2CO3 = (2*22.99) + 12.01 + (3*16.00) = 105.99 g/mol

2. Calculate the number of moles of sodium carbonate: - Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3 - Moles of Na2CO3 = 21.2 g / 105.99 g/mol = 0.2 moles

3. Use the mole ratio from the balanced equation to find the moles of calcium carbonate formed: - Moles of CaCO3 = Moles of Na2CO3

4. Calculate the mass of calcium carbonate formed: - Mass of CaCO3 = Moles of CaCO3 * Molar mass of CaCO3 - Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3*16.00 g/mol (O) = 100.09 g/mol - Mass of CaCO3 = 0.2 moles * 100.09 g/mol = 20.02 g

The mass of the precipitate (calcium carbonate) formed is approximately 20.02 grams.