# Lead ions can be precipitated from solution with KCl according to the following reaction:Pb2+(aq)+2KCl(aq)→PbCl2(s)+2K+(aq) When 28.6 g KCl is added to a solution containing 25.5 gPb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.0 g . Determine the limiting reactant.

Question
144 views

Lead ions can be precipitated from solution with KCl according to the following reaction:

Pb2+(aq)+2KCl(aq)→PbCl2(s)+2K+(aq)
When 28.6 g KCl is added to a solution containing 25.5 gPb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.0 g .

Determine the limiting reactant.

check_circle

Step 1

Given information:

Mass of KCl = 28.6 g

Mass of lead ions (Pb2+) = 25.5 g

Mass of PbCl2 = 29.0 g

Step 2

Number of moles of a compound can be expressed in terms of given mass of the compound and molar mass. The expression showing relation between mass and molar mass of a compound is represented as follows:

Step 3

Thus, the number of moles contained in 28.6 g of KCl and 25.5 ...

### Want to see the full answer?

See Solution

#### Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in