Question

К 137 мл раствора сульфата натрия с массовой долей 0,12 и плотностью 1,11 г/мл прилили 292 мл раствора хлорида бария с массовой долей 0,08 и плотностью 1,08 г/мл. Определите массу образовавшегося осадка.

Answer 1

Na2SO4+BaCl2→2NaCl+BaSO4↓
m р-ра(Na2SO4)=Vр-ра•плотность=137мл•1,11г/мл=152,07г в растворе
m(Na2SO4)=152,07•0,12=18,25г
m р-ра(BaCl2)=292мл•1,08г/мл=315,36 в растворе
m(BaCl2)=315,36•0,08=25,23г
Количество вещества (Na2SO4)=0,12=количеству вещества (BaCl2)
m осадка =0,12•233=27,96г
Ответ:27,96г

Answer 2

Calculation of the Mass of Precipitate Formed

To determine the mass of the precipitate formed, we need to calculate the amount of each substance involved and then use stoichiometry to find the mass of the precipitate.

Given information: - Volume of sodium sulfate solution (V1) = 137 mL - Mass fraction of sodium sulfate (w1) = 0.12 - Density of sodium sulfate solution (d1) = 1.11 g/mL - Volume of barium chloride solution (V2) = 292 mL - Mass fraction of barium chloride (w2) = 0.08 - Density of barium chloride solution (d2) = 1.08 g/mL

To calculate the mass of the precipitate, we need to find the moles of sodium sulfate (Na2SO4) and barium chloride (BaCl2) present in the solutions.

Moles of Sodium Sulfate (Na2SO4)

The moles of sodium sulfate can be calculated using the formula:

moles = mass / molar mass

The molar mass of sodium sulfate (Na2SO4) is calculated as follows: - Molar mass of Na = 22.99 g/mol - Molar mass of S = 32.07 g/mol - Molar mass of O = 16.00 g/mol

Molar mass of Na2SO4 = (2 * 22.99) + 32.07 + (4 * 16.00) = 142.04 g/mol

The mass of sodium sulfate (m1) can be calculated as follows: - m1 = V1 * d1

Substituting the given values: - m1 = 137 mL * 1.11 g/mL = 151.07 g

The moles of sodium sulfate (n1) can be calculated as follows: - n1 = m1 / Molar mass of Na2SO4

Substituting the calculated values: - n1 = 151.07 g / 142.04 g/mol = 1.063 mol

Moles of Barium Chloride (BaCl2)

The moles of barium chloride can be calculated using the same method as above.

The molar mass of barium chloride (BaCl2) is calculated as follows: - Molar mass of Ba = 137.33 g/mol - Molar mass of Cl = 35.45 g/mol

Molar mass of BaCl2 = 137.33 + (2 * 35.45) = 208.23 g/mol

The mass of barium chloride (m2) can be calculated as follows: - m2 = V2 * d2

Substituting the given values: - m2 = 292 mL * 1.08 g/mL = 315.36 g

The moles of barium chloride (n2) can be calculated as follows: - n2 = m2 / Molar mass of BaCl2

Substituting the calculated values: - n2 = 315.36 g / 208.23 g/mol = 1.514 mol

Stoichiometry and Mass of Precipitate

The balanced chemical equation for the reaction between sodium sulfate and barium chloride is:

Na2SO4 + BaCl2 -> 2NaCl + BaSO4

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

Since the stoichiometric ratio is 1:1, the moles of barium sulfate formed (n3) will be equal to the moles of sodium sulfate (n1) or barium chloride (n2) used.

The mass of the precipitate (m3) can be calculated as follows: - m3 = n3 * Molar mass of BaSO4

Substituting the calculated values: - m3 = 1.063 mol * (137.33 + 32.07 + (4 * 16.00)) g/mol = 208.52 g

Therefore, the mass of the precipitate formed is 208.52 grams.

Note: The given search results did not provide direct information for this specific calculation. The calculation is based on the principles of stoichiometry and molar mass calculations.