Question

К раствору хлорида железа(III) массой 488 г и массовой долей 10% прилили избыток раствора нитрата серебра. Вычислите массу образовавшегося осадка.

Answer 1

Vot pojaluysta ) udachi ☺️

Answer 2

Calculation of the Mass of the Precipitate Formed

To calculate the mass of the precipitate formed when an excess of silver nitrate solution is added to a solution of iron(III) chloride, we need to consider the balanced chemical equation for the reaction between iron(III) chloride and silver nitrate.

The balanced chemical equation for the reaction is as follows:

FeCl3 + 3AgNO3 → Fe(NO3)3 + 3AgCl

From the equation, we can see that one mole of iron(III) chloride reacts with three moles of silver nitrate to produce one mole of iron(III) nitrate and three moles of silver chloride.

To calculate the mass of the precipitate formed, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

In this case, the iron(III) chloride is the limiting reactant because it is present in a specific mass and the silver nitrate is added in excess.

To calculate the mass of the precipitate, we need to determine the number of moles of iron(III) chloride and then use the stoichiometry of the balanced equation to find the number of moles of silver chloride formed. Finally, we can convert the moles of silver chloride to grams using its molar mass.

Given: - Mass of iron(III) chloride = 488 g - Mass fraction of iron(III) chloride = 10%

To calculate the number of moles of iron(III) chloride, we can use its molar mass. The molar mass of iron(III) chloride (FeCl3) can be calculated by adding the atomic masses of iron (Fe) and three times the atomic mass of chlorine (Cl).

The atomic mass of iron (Fe) is approximately 55.845 g/mol, and the atomic mass of chlorine (Cl) is approximately 35.453 g/mol.

Molar mass of FeCl3 = (1 * atomic mass of Fe) + (3 * atomic mass of Cl) = (1 * 55.845 g/mol) + (3 * 35.453 g/mol) = 162.204 g/mol

To calculate the number of moles of iron(III) chloride, we can use the formula:

Number of moles = Mass / Molar mass

Number of moles of FeCl3 = 488 g / 162.204 g/mol ≈ 3.01 mol

According to the balanced chemical equation, one mole of iron(III) chloride reacts with three moles of silver nitrate to produce three moles of silver chloride.

Therefore, the number of moles of silver chloride formed will be three times the number of moles of iron(III) chloride.

Number of moles of AgCl = 3 * Number of moles of FeCl3 ≈ 3 * 3.01 mol ≈ 9.03 mol

To calculate the mass of the precipitate (silver chloride), we can use its molar mass. The molar mass of silver chloride (AgCl) can be calculated by adding the atomic masses of silver (Ag) and chlorine (Cl).

The atomic mass of silver (Ag) is approximately 107.8682 g/mol, and the atomic mass of chlorine (Cl) is approximately 35.453 g/mol.

Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl = 107.8682 g/mol + 35.453 g/mol = 143.3212 g/mol

To calculate the mass of the precipitate, we can use the formula:

Mass = Number of moles * Molar mass

Mass of AgCl = 9.03 mol * 143.3212 g/mol ≈ 1294.7 g

Therefore, the mass of the precipitate (silver chloride) formed is approximately 1294.7 grams.

Please note that the calculations provided are based on the given information and assumptions.