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
Asked Oct 28, 2019
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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.

 

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Expert Answer

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:

Givenmass
Number of moles
Molar mass
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Givenmass Number of moles Molar mass

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Step 3

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

28.6g
74.5513g.mol
0.384 mol
25.5g
207.2g.mol
=0.123mol
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28.6g 74.5513g.mol 0.384 mol 25.5g 207.2g.mol =0.123mol

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