Question

Какая масса осадка образуется при взаимодействии избытка раствора хлорида кальция с 65,6 грамм раствора фосфата натрия с массовой долей растворенного вещества 10%?

Answer 1

3CaCl2 + 2Na3PO4 = Ca3(PO4)2 + 6NaCl
65.6гр *10%/100%=6,56гр
6,56 --- х=6,2                        328- это масса 2Na3PO4
328 ---310                             310- это масса осадка Ca3(PO4)2

Answer 2

Calculation of Precipitate Mass

To calculate the mass of the precipitate formed when an excess of calcium chloride solution reacts with 65.6 grams of sodium phosphate solution with a mass fraction of dissolved substance of 10%, we can use the stoichiometry of the reaction and the principles of chemical reactions.

The balanced chemical equation for the reaction between calcium chloride and sodium phosphate is:

3CaCl2 + 2Na3PO4 → Ca3(PO4)2 + 6NaCl

From the balanced equation, we can see that 3 moles of calcium chloride react with 2 moles of sodium phosphate to produce 1 mole of calcium phosphate.

Using the given mass of the sodium phosphate solution and the mass fraction of the dissolved substance, we can calculate the mass of the dissolved substance in the solution:

Mass of dissolved substance = Mass of solution × Mass fraction of dissolved substance Mass of dissolved substance = 65.6 g × 0.10 = 6.56 g

Now, we can use the stoichiometry of the reaction to find the limiting reactant and the mass of the precipitate formed.

Let's calculate the mass of the precipitate formed using the given information.

The mass of the precipitate formed can be calculated using the stoichiometry of the reaction and the molar masses of the compounds involved.

Calculation Steps:

1. Calculate the moles of the dissolved substance in the sodium phosphate solution. 2. Determine the limiting reactant. 3. Calculate the moles of the precipitate formed. 4. Convert the moles of the precipitate to mass.

Let's proceed with the calculations.