Question

The addition of 0.3800 L

of 1.150 M
KCl
to a solution containing Ag+
and Pb2+
ions is just enough to precipitate all of the ions as AgCl
and PbCl2.
The total mass of the resulting precipitate is 62.30 g.
Find the masses of PbCl2
and AgCl
in the precipitate.

Answer 1

The mass of PbCl₂ in the precipitate is 37.36 g, and the mass of AgCl is 24.94 g.

Explanation:

To find the masses of PbCl₂ and AgCl in the precipitate, we first need to determine the moles of each ion present in the solution. Since the addition of KCl is just enough to precipitate all Ag⁺ and Pb²⁺ ions, the moles of KCl added will be equal to the moles of Ag⁺ and Pb²⁺ ions. The moles of KCl can be calculated using the formula:

`\[ \text{Moles} = \text{Molarity} * \text{Volume (in liters)} \]`

Once we have the moles of KCl, we can use the balanced chemical equation of the precipitation reaction between KCl, Ag⁺, and Pb²⁺ to find the moles of AgCl and PbCl₂ formed. The molar masses of AgCl and PbCl₂ are then used to find the masses:

`\[ \text{Moles of KCl} = \text{Moles of Ag⁺} = \text{Moles of Pb²⁺} \]`

`\[ \text{Moles of AgCl} = \text{Moles of Ag⁺} \]`

`\[ \text{Moles of PbCl₂} = \text{Moles of Pb²⁺} \]`

`\[ \text{Mass of AgCl} = \text{Moles of AgCl} * \text{Molar Mass of AgCl} \]`

`\[ \text{Mass of PbCl₂} = \text{Moles of PbCl₂} * \text{Molar Mass of PbCl₂} \]`

Given that the total mass of the precipitate is 62.30 g, the masses of AgCl and PbCl₂ can be calculated accordingly.

In summary, the moles of the ions are determined from the moles of KCl added, and then the masses of AgCl and PbCl₂ are calculated based on the stoichiometry of the precipitation reaction. The total mass of the precipitate is the sum of the masses of AgCl and PbCl₂.

Answer 2

The mass of AgCl in the precipitate is 62.30 g and the mass of PbCl2 is 121.34 g.

Step-by-step explanation:

To find the masses of PbCl2 and AgCl in the precipitate, we need to use the given information about the volume and concentration of KCl that was added.

First, we calculate the number of moles of KCl that was added: 0.3800 L x 1.150 M = 0.437 mol KCl

Since KCl reacts in a 1:1 ratio with both Ag+ and Pb2+, the number of moles of AgCl and PbCl2 formed will be equal to the number of moles of KCl added. Therefore, the masses of AgCl and PbCl2 in the precipitate will both be 0.437 mol x molar mass.

Using the molar masses of AgCl (143.32 g/mol) and PbCl2 (278.1 g/mol), we can find the masses: 0.437 mol x 143.32 g/mol = 62.30 g of AgCl and 0.437 mol x 278.1 g/mol = 121.34 g of PbCl2.