 Chemistry
12 June, 06:41

# Phosgenite, Pb2Cl2CO3, is prepared by the reaction shown. If 20.0 g if lead (11) oxide and 20.0 g of sodiumchloride are reacted in the presence of excess water and carbon dioxide, 12.81 g of phosgenite is produced.Calculate the percent yield for the reaction.PbO (s) + NaCl (aq) + H20 (1) + CO2 (g) → Pb2Cl2CO3 (s) + NaOH (aq).

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1. 12 June, 07:46
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Percent yield: 52.6 %

Explanation:

1) Unbalanced chemical equation (given)

PbO (s) + NaCl (aq) + H₂O (l) + CO₂ (g) → Pb₂Cl₂CO₃ (s) + NaOH (aq).

2) Balanced chemical equation

Add the coefficients that make the number of atoms of each element on the left side equal the number of atoms of the same element in the right side:

2PbO (s) + 2NaCl (aq) + H₂O (l) + CO₂ (g) → Pb₂Cl₂CO₃ (s) + 2NaOH (aq).

3) Write the mole ratios:

2 mol PbO : 2 mol NaCl : 1 molH₂O : CO₂ : 1 mol Pb₂Cl₂CO₃ (s) : 2 mol NaOH

You are only interested in PbO, NaCl, and Pb₂Cl₂CO₃

4) Convert the masses of PbO, NaCl, and Pb₂Cl₂CO₃ in number of moles.

number of moles = mass in grams / molar mass

Molar masses

PbO: 223.2 g/mol NaCl: 58.44 g/mol Pb₂Cl₂CO₃: 543.3 g/mol

Moles

PbO: 20.0 g / 223.2 g/mol = 0.08961 mol NaCl: 20.0 g / 58.44 g/mol = 0.3422 mol Pb₂Cl₂CO₃: 12.81 g / 543.3 g/mol = 0.023578 mol

5) Limiting reactant

As per the theoretical mole ratios, PbO and NaCl react in a ratio 2: 2, i. e. 1:1.

Then, 0.08961 mol of PbO will react completely with the same number of moles of NaCl, and 0.3422 mol - 0.08961 will be left.

6) Calculate the theoretical yield

Using the limiting reactant calculate how much would be produced

Stoichiometric proportion:

2 mol PbO / 1 mol Pb₂Cl₂CO₃ = 0.08961 mol PbO / x

Solve for x:

x = 0.08961 * 1 / 2 mol Pb₂Cl₂CO₃ = 0.044805 mol Pb₂Cl₂CO₃

Convert to mass:

mass = number of moles * molar mass = 0.044805 mol * 543.3 g/mol = 24.34 g

7) Percent yield (%)

% = (actual yield / theoretical yield) * 100

% = (12.81 g / 24.34 g) * 100 = 52.63%