Question

Lead ion can be precipitated from solution with NaCl according to the reaction: Pb2 (aq)+2NaCl(aq) PbCl2(s)+ 2Na (aq) When 135.8 g of NaCI are added to a solution containing 195.7g of Pb2*, a PbCl2 precipitate forms. The precipice is filtered and dried ad found to have a mass of 252.4g. Determine the limiting reactant, theoretical yield of PbC12, and percent yield for the reaction

Answer 1

Answer:

$Pb^{2+}$ is limiting reagent.

Theoretical yield = 262.67 g

% yield = (Experimental yield / Theoretical yield) × 100 = (252.4 / 262.67) × 100 = 96.1 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $NaCl$

Given mass = 135.8 g

Molar mass of $NaCl$ = 58.44 g/mol

Moles of $NaCl$ = 135.8 g / 58.44 g/mol = 2.3238 moles

Given: For $Pb^{2+}$

Given mass = 195.7 g

Molar mass of $Pb^{2+}$ = 207.2 g/mol

Moles of $Pb^{2+}$ = 195.7 g / 207.2 g/mol = 0.9445 moles

According to the given reaction:

$Pb^{2+}_{(aq)}+2NaCl_{(aq)}\rightarrow PbCl_2_{(s)}+2Na^+_{(aq)}$

1 mole of $Pb^{2+}$ react with 2 moles of $NaCl$

Also,

0.9445 mole of $Pb^{2+}$ react with 2*0.9445 moles of $NaCl$

Moles of $NaCl$ = 1.889 moles

Available moles of $NaCl$ = 2.3238 moles  (Extra)

Limiting reagent is the one which is present in small amount. Thus, $Pb^{2+}$ is limiting reagent.

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

1 mole of $Pb^{2+}$ gives 1 mole of $PbCl_2$

0.9445 mole of $Pb^{2+}$ gives 0.9445 mole of $PbCl_2$

Mole of $PbCl_2$  = 0.9445 moles

Molar mass of $PbCl_2$ = 278.1 g/mol

Mass of $PbCl_2$ = Moles × Molar mass = 0.9445 × 278.1 g = 262.67 g

Theoretical yield = 262.67 g

Given experimental yield = 252.4 g

% yield = (Experimental yield / Theoretical yield) × 100 = (252.4 / 262.67) × 100 = 96.1 %