Question

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2 KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2 KNO3(aq) The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

$Moles=1.20 \times {25.0\times 10^{-3}}\ moles$

Moles of potassium chloride  = 0.03 moles

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=0.900 \times {15.0\times 10^{-3}}\ moles$

Moles of lead(II) nitrate  = 0.0135 moles

According to the given reaction:

$2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

Theoretical yield = 3.75435 g

Given experimental yield = 2.45 g

% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %