Question

Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. $$\mathrm{PbO}(s)+\mathrm{C}(s) \rightarrow \mathrm{Pb}(l)+\mathrm{CO}(g)$$ Calculate the expected yield of lead if \(50.0 \mathrm{kg}\) of lead oxide is heated with \(50.0 \mathrm{kg}\) of carbon.

Short answer

The expected yield of lead is approximately 46.37 kg when 50.0kg of lead oxide is heated with 50.0kg of carbon.

Step-by-step solution

Calculate molar mass of each compound

First, we need to find the molar mass of each compound in the balanced chemical equation. The molar mass of a compound is the sum of the atomic masses of its constituent elements, multiplied by their respective moles. For PbO (Lead(II) oxide): Pb = 207.2 g/mol O = 16.0 g/mol Molar mass of PbO = 207.2 + 16.0 = 223.2 g/mol For C (Carbon): C = 12.0 g/mol Molar mass of C = 12.0 g/mol For Pb (Lead): Molar mass of Pb = 207.2 g/mol For CO (Carbon monoxide): C = 12.0 g/mol O = 16.0 g/mol Molar mass of CO = 12.0 + 16.0 = 28.0 g/mol

Calculate moles of each reactant

Now, convert the mass of each reactant (Lead(II) oxide and Carbon) to moles by dividing their mass by their molar mass. Moles of PbO = (50,000 g) / (223.2 g/mol) = 223.88 mol Moles of C = (50,000 g) / (12.0 g/mol) = 4166.67 mol

Determine the limiting reactant

Now, we need to determine the limiting reactant by comparing the mole ratio of the given reactants. The balanced chemical equation tells us: 1 mol PbO + 1 mol C → 1 mol Pb + 1 mol CO Divide the moles of each reactant by their stoichiometric coefficients: PbO: 223.88 mol / 1 = 223.88 C: 4166.67 mol / 1 = 4166.67 Since the ratio for PbO is smaller, PbO is the limiting reactant.

Calculate the expected yield of Lead

Using stoichiometry and the limiting reactant (PbO), we can now calculate the expected yield of Lead. From the balanced chemical equation, we know that: 1 mol PbO → 1 mol Pb Moles of Pb = Moles of PbO = 223.88 mol Now, we'll convert moles of Pb to mass: Mass of Pb = Moles of Pb × Molar mass of Pb = 223.88 mol × 207.2 g/mol = 46,374.34 g Thus, the expected yield of lead is 46,374.34 g or approximately 46.37 kg.

Key concepts

Molar Mass Calculation

Understanding the molar mass of compounds is an essential step in mastering chemical stoichiometry. To calculate the molar mass, you need to sum up the atomic masses of all the atoms that comprise a molecule. It acts as a bridge between the mass of a substance and the amount of moles present.

Let's delve into an example. For Lead(II) oxide (PbO), the molar mass is computed by adding the atomic mass of lead (207.2 g/mol) to the atomic mass of oxygen (16.0 g/mol), resulting in a molar mass of PbO equalling 223.2 g/mol. Similarly, calculate the molar mass of any compound by summing up the atomic masses of its constituent elements, each multiplied by the number of times the element occurs in the compound.

When you're given a certain mass of a substance, you can convert it to moles by dividing the mass by the molar mass. This step is crucial as it sets the stage for further stoichiometric calculations, like determining the limiting reactant or computing theoretical yields.

Limiting Reactant Determination

The limiting reactant in a chemical reaction is the reactant that runs out first, thus determining the maximum amount of product that can be formed. To find the limiting reactant, one should compare the mole ratios of the reactants to their coefficients in the balanced chemical equation.

In our exercise, the balanced equation is as follows: \(\mathrm{PbO}(s) + \mathrm{C}(s) \rightarrow \mathrm{Pb}(l) + \mathrm{CO}(g)\). With the molar amounts of PbO and C calculated, you then compare these amounts to their stoichiometric coefficients. In this case, both PbO and C have a coefficient of 1, indicating a 1:1 ratio. Whichever reactant has the smaller mole ratio is the limiting reactant.

In practice, after converting the mass of each reactant to moles, you divide the moles of each reactant by their respective coefficients. The smallest result indicates the limiting reactant; this is PbO in our example. Identifying the limiting reactant is crucial as it impacts the amount of products formed, a foundation for yield predictions in chemical reactions.

Theoretical Yield Computation

The theoretical yield is the maximum amount of product that can be generated from a chemical reaction under perfect conditions. It is calculated based on the limiting reactant since it is the reactant that will be completely consumed first during the reaction.

To compute the theoretical yield, you convert the moles of the limiting reactant to moles of the desired product using the mole-to-mole ratio from the balanced equation. Then, multiply the moles of the product by its molar mass to find the mass of the product, which is the theoretical yield.

In the provided example, 1 mole of PbO produces 1 mole of Pb. Since PbO is the limiting reactant with 223.88 moles available, it can produce the same number of moles of Pb. Multiplying the moles of Pb by its molar mass (207.2 g/mol) gives the mass of Pb in grams, which can be converted to kilograms. The result is the theoretical yield of the reaction. This process is essential for both practical laboratory work and industrial applications where predicting the amount of product is vital for economic and efficiency considerations.