Question

К 300 г 6%-ного раствора сульфата магния прилили избыток раствора фосфата натрия. Какова масса выпавшего осадка?

Answer 1

Вроде так, но если есть возможность, проверь

Answer 2

Calculating the Mass of Precipitate

To calculate the mass of the precipitate formed when an excess of sodium phosphate solution is added to 300g of a 6% magnesium sulfate solution, we can use the principles of stoichiometry and chemical reactions.

The balanced chemical equation for the reaction between magnesium sulfate and sodium phosphate is: MgSO4 + Na3PO4 → Mg3(PO4)2 + Na2SO4

From the balanced chemical equation, we can see that one mole of magnesium sulfate reacts with one mole of sodium phosphate to produce one mole of magnesium phosphate. Using this information, we can calculate the mass of the precipitate formed.

Calculation Steps

1. Calculate the number of moles of magnesium sulfate in 300g of 6% solution. 2. Use the stoichiometry of the reaction to find the number of moles of magnesium phosphate formed. 3. Convert the moles of magnesium phosphate to grams to find the mass of the precipitate.

Calculation

1. Calculate the number of moles of magnesium sulfate: - The molar mass of magnesium sulfate (MgSO4) is 120.37 g/mol. - The percentage concentration of the solution is 6%, so the mass of magnesium sulfate in 300g of the solution is 0.06 * 300g = 18g. - The number of moles of magnesium sulfate is given by: moles = mass / molar mass = 18g / 120.37 g/mol ≈ 0.1497 moles.

2. Use stoichiometry to find the moles of magnesium phosphate formed: - From the balanced chemical equation, 1 mole of magnesium sulfate reacts with 1 mole of sodium phosphate to produce 1 mole of magnesium phosphate. - Therefore, the number of moles of magnesium phosphate formed is also approximately 0.1497 moles.

3. Convert moles of magnesium phosphate to grams: - The molar mass of magnesium phosphate (Mg3(PO4)2) is 262.86 g/mol. - The mass of the precipitate formed is given by: mass = moles * molar mass = 0.1497 moles * 262.86 g/mol ≈ 39.39g.

Answer

The mass of the precipitate formed when an excess of sodium phosphate solution is added to 300g of a 6% magnesium sulfate solution is approximately 39.39 grams.