A25.0-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.

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

Or,

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 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 of lead(II) nitrate  = 0.0135 moles

According to the given reaction:

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 %

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