Question

Lead(II) carbonate, also called "white lead," was formerly used as a pigment in white paints. However, because of its toxicity, lead can no longer be used in paints intended for residential homes. Lead(II) carbonate is prepared industrially by reaction of aqueous lead(II) acetate with carbon dioxide gas. The unbalanced equation is $$ \mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{PbCO}_{3}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) $$ Suppose an aqueous solution containing 1.25 g of lead(II) acctate is treated with 5.95 g of carbon dioxide. Calculate the theoretical yield of lead carbonate.

Short answer

The theoretical yield of lead(II) carbonate, when reacting 1.25 g of lead(II) acetate with 5.95 g of carbon dioxide, is 1.03 g.

Step-by-step solution

Balance the chemical equation

First, let's balance the given chemical equation: \[ \mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{PbCO}_{3}(s)+2\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \]

Convert masses to moles

Now, we need to convert the given masses of lead(II) acetate (1.25 g) and carbon dioxide (5.95 g) to moles. We will use their molar masses: Molar mass of Pb(C2H3O2)2 = 325.29 g/mol Molar mass of CO2 = 44.01 g/mol Moles of Pb(C2H3O2)2 = (1.25 g) / (325.29 g/mol) = 0.00384 mol Moles of CO2 = (5.95 g) / (44.01 g/mol) = 0.135 mol

Determine the limiting reactant

Using the balanced equation, we can see that the mole ratio between Pb(C2H3O2)2 and CO2 is 1:1. Divide the moles of each by their respective stoichiometric coefficients to determine the limiting reactant: For Pb(C2H3O2)2: 0.00384 mol / 1 = 0.00384 For CO2: 0.135 mol / 1 = 0.135 Since 0.00384 is smaller than 0.135, the limiting reactant is Pb(C2H3O2)2.

Calculate the theoretical yield of lead(II) carbonate

Now, we will use the stoichiometry and the limiting reactant to calculate the theoretical yield of PbCO3: Moles of PbCO3 = moles of limiting reactant (Pb(C2H3O2)2) = 0.00384 mol Next, we will convert moles of PbCO3 to grams using its molar mass: Molar mass of PbCO3 = 267.21 g/mol Theoretical yield of PbCO3 = (0.00384 mol) × (267.21 g/mol) = 1.03 g So, the theoretical yield of lead(II) carbonate is 1.03 g.

Key concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is entirely consumed first, limiting the amount of products formed. To find which reactant is limiting, you first need a balanced chemical equation. Then, convert the initial masses of reactants into moles. Compare these moles with the stoichiometric ratios from the balanced equation.

For instance, in our exercise, we had lead(II) acetate and carbon dioxide reacting. We discovered that lead(II) acetate was the limiting reactant because it had fewer moles available compared to what was required by the balanced chemical equation. Therefore, no matter how much carbon dioxide is present, the reaction cannot produce more product once all the lead(II) acetate has been used up.

• Identify the limiting reactant by comparing moles available to required stoichiometric amounts.
• The limiting reactant determines the maximum amount of product that can be formed.

Chemical Equation Balancing

Balancing a chemical equation is crucial because it ensures that matter is conserved during the reaction. Each element must have the same number of atoms on both sides of the equation.

In the provided exercise, we started with an unbalanced equation. By adjusting the coefficients, we ended up with: \[ \mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{PbCO}_{3}(s)+2\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \]This balanced equation reflects equal numbers of atoms for each element, ensuring that the reaction adheres to the laws of conservation of mass.

• Check each element and adjust coefficients to balance the equation.
• Ensure the total mass of reactants equals the total mass of products.

Theoretical Yield

The theoretical yield of a reaction is the amount of product expected if the reaction proceeds perfectly, consuming all of the limiting reactant. This can be calculated once the limiting reactant is known.

In our exercise, knowing that lead(II) acetate was the limiting reactant allowed us to predict the maximum amount of lead(II) carbonate that could potentially be produced. We calculated this using the converted moles of the limiting reactant and multiplied by the molar mass of the desired product.

• Theoretical yield is always calculated in ideal conditions without losses.
• It is an important benchmark to measure the efficiency of the actual reaction.

Molar Mass Calculation

Molar mass is essential for converting between the mass of a substance and the amount in moles. It takes into account the atomic masses of all elements in a compound.

For example, lead(II) acetate has a molar mass of 325.29 g/mol and carbon dioxide is 44.01 g/mol. These values were used to convert the mass of each reactant into moles, a necessary step for determining the limiting reactant and calculating the theoretical yield.

• Find the molecular weight by adding up the atomic masses of the elements in the compound.
• Use molar mass to convert mass to moles: \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).
• In stoichiometry, always express quantities in moles for accurate calculations.