Question

Вычислите массу осадка образующегося в результате реакций 5,5г CLK с избыдком карбоатом NA

Answer 1

KCl не реагирует с Na2CO3, так как установится равновесие 2KCl+Na2CO3=2NaCl+K2CO3

Answer 2

Calculating the Mass of Precipitate Formed

To calculate the mass of the precipitate formed in a chemical reaction, we need to first determine the balanced chemical equation for the reaction between 5.5g of calcium chloride (CaCl2) and excess sodium carbonate (Na2CO3).

The balanced chemical equation for the reaction between calcium chloride and sodium carbonate is as follows: ``` CaCl2 + Na2CO3 -> CaCO3 + 2NaCl ```

From the balanced chemical equation, we can see that 1 mole of calcium chloride (CaCl2) reacts with 1 mole of sodium carbonate (Na2CO3) to produce 1 mole of calcium carbonate (CaCO3) and 2 moles of sodium chloride (NaCl).

Calculating the Moles of Calcium Chloride

To calculate the moles of calcium chloride, we can use the formula: ``` moles = mass / molar mass ``` The molar mass of calcium chloride (CaCl2) is 40.08 g/mol (calcium) + 2 * 35.45 g/mol (chlorine) = 110.98 g/mol.

So, the moles of calcium chloride can be calculated as: ``` moles = 5.5g / 110.98 g/mol ≈ 0.0495 moles ```

Calculating the Limiting Reactant

To determine the limiting reactant, we compare the moles of calcium chloride to the moles of sodium carbonate. From the balanced chemical equation, we know that the stoichiometric ratio of calcium chloride to sodium carbonate is 1:1.

Since the moles of calcium chloride are approximately 0.0495, we also need to calculate the moles of sodium carbonate.

Calculating the Moles of Sodium Carbonate

The molar mass of sodium carbonate (Na2CO3) is 22.99 g/mol (sodium) + 2 * 16.00 g/mol (carbon) + 3 * 16.00 g/mol (oxygen) = 105.99 g/mol.

So, the moles of sodium carbonate can be calculated as: ``` moles = 5.5g / 105.99 g/mol ≈ 0.052 moles ```

Determining the Limiting Reactant

Since the moles of calcium chloride and sodium carbonate are approximately 0.0495 and 0.052 respectively, we can see that calcium chloride is the limiting reactant because it produces fewer moles of product compared to sodium carbonate.

Calculating the Mass of Precipitate Formed

The balanced chemical equation shows that 1 mole of calcium chloride produces 1 mole of calcium carbonate. The molar mass of calcium carbonate (CaCO3) is 40.08 g/mol (calcium) + 12.01 g/mol (carbon) + 3 * 16.00 g/mol (oxygen) = 100.09 g/mol.

So, the mass of the precipitate (calcium carbonate) formed can be calculated as: ``` mass = moles of calcium chloride * molar mass of calcium carbonate ≈ 0.0495 moles * 100.09 g/mol ≈ 4.96g ```

The mass of the precipitate (calcium carbonate) formed in the reaction is approximately 4.96 grams.