Question

The addition of 0.275 L of 1.62 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 62.37 g. Find the mass of PbCl, and AgCl in the precipitate.

Answer 1

The mass of AgCl in the precipitate is 63.98 g, and the mass of PbCl2 in the precipitate is 124.42 g.

Step-by-step explanation:

To find the mass of PbCl2 and AgCl in the precipitate, we need to use stoichiometry and the concept of limiting reagents. First, let's calculate the moles of KCl that were added to the solution:

Moles of KCl = volume of KCl x molarity of KCl = 0.275 L x 1.62 M = 0.4465 moles

Since the reaction is 1:1 for KCl and AgCl, this means that 0.4465 moles of AgCl were formed. The molar mass of AgCl is 143.32 g/mol, so the mass of AgCl in the precipitate is:

Mass of AgCl = moles of AgCl x molar mass of AgCl = 0.4465 moles x 143.32 g/mol = 63.98 g

Similarly, for PbCl2, the reaction is 1:1 for KCl and PbCl2. Therefore, 0.4465 moles of PbCl2 were formed. The molar mass of PbCl2 is 278.1 g/mol, so the mass of PbCl2 in the precipitate is:

Mass of PbCl2 = moles of PbCl2 x molar mass of PbCl2 = 0.4465 moles x 278.1 g/mol = 124.42 g

Answer 2

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Step-by-step explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,

x as 0.115 and y as 0.165

Now the mass of AgCl will be,

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,

278.1 × 0.165 = 45.88 grams.