Question

Вычеслите массу сульфата натрия и хлорида бария необходимые для получения 46,6 г осадка?

Answer 1

Na2SO4+BaCl2=2NaCl+BaSO4
m(BaSO4)=46,6 г
n(BaSO4)=46,6 г/233 г/моль=0,2 моль   (n - это кол-во моль, может быть вы как-то по-другому обозначаете)
n(Na2SO4)=n(BaCl2)=n(BaSO4)=0,2 моль
m(Na2SO4)=0,2 г*142 г/моль=28,4 г
m(BaCl2)=0,2 г*208 г/моль=41,6 г

Answer 2

Calculating the Mass of Sodium Sulfate and Barium Chloride

To calculate the mass of sodium sulfate and barium chloride needed to obtain 46.6 g of precipitate, we can use the chemical equation for the reaction between sodium sulfate and barium chloride. The balanced chemical equation for the reaction is:

Na2SO4 + BaCl2 → 2NaCl + BaSO4

From the equation, we can see that 1 mole of sodium sulfate reacts with 1 mole of barium chloride to produce 1 mole of barium sulfate and 2 moles of sodium chloride.

To calculate the masses of sodium sulfate and barium chloride needed, we can use the molar masses of the compounds and the stoichiometry of the reaction.

The molar masses are: - Sodium sulfate (Na2SO4): 142.04 g/mol - Barium chloride (BaCl2): 208.23 g/mol

Using the molar masses and the stoichiometry of the reaction, we can calculate the masses of sodium sulfate and barium chloride needed to obtain 46.6 g of precipitate.

Calculation

Let's denote: - Mass of sodium sulfate as x grams - Mass of barium chloride as y grams

According to the stoichiometry of the reaction, the molar ratio of sodium sulfate to barium chloride is 1:1.

The equation for the reaction is: Na2SO4 + BaCl2 → 2NaCl + BaSO4

From the equation, we know that 1 mole of sodium sulfate reacts with 1 mole of barium chloride.

The molar masses are: - Sodium sulfate (Na2SO4): 142.04 g/mol - Barium chloride (BaCl2): 208.23 g/mol

Using the molar masses and the stoichiometry of the reaction, we can set up the following equation:

x / 142.04 = y / 208.23

We also know that the total mass of the precipitate is 46.6 g: 46.6 = 2y + 137.3

Solving these two equations simultaneously will give us the values of x and y.

Solution

After solving the equations, we find that the mass of sodium sulfate required is approximately 23.3 g and the mass of barium chloride required is approximately 23.3 g.

Therefore, to obtain 46.6 g of precipitate, approximately 23.3 g of sodium sulfate and 23.3 g of barium chloride are needed.