Question

Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is: $$\mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})$$ a. Balance the equation, and determine the theoretical yield of PbO if 200.0 g of PbS is heated. b. What is the percent yield if 170.0 g of PbO is obtained?

Short answer

The balanced equation for the reaction is \(2 \mathrm{PbS}(s) + 3 \mathrm{O_{2}}(g) \rightarrow 2 \mathrm{PbO}(s) + 2 \mathrm{SO_{2}}(g)\). The theoretical yield of PbO when 200.0 g of PbS is heated is 186.5 g, and the percent yield for the reaction, given 170.0 g of PbO obtained, is approximately 91.1%.

Step-by-step solution

1. Balance the chemical equation

To balance the provided equation, we will assign stoichiometric coefficients to the chemical species on both sides of the equation until they are balanced. \(2 \mathrm{PbS}(s) + 3 \mathrm{O_{2}}(g) \rightarrow 2 \mathrm{PbO}(s) + 2 \mathrm{SO_{2}}(g)\) Now that we have the balanced equation, we can proceed to find the theoretical yield of \(PbO\).

2. Calculate the theoretical yield of PbO

To find the theoretical yield of PbO, we will use the balanced chemical equation and the molar masses of the substances involved. First, we need to calculate the number of moles of lead(II) sulfide (PbS) in the given sample. Given mass: \(200.0 \ g\) Molar mass of PbS: \((207.2 + 32.1) \ g/mol = 239.3 \ g/mol\) Number of moles of PbS: \(\frac{200.0}{239.3} = 0.836 \ mol\) Now, using the stoichiometry of the balanced equation (2 moles of PbS produce 2 moles of PbO), we can find the number of moles of PbO produced. Number of moles of PbO: \(0.836 \ mol\) Finally, we convert the number of moles of PbO back to grams to find the theoretical yield. Molar mass of PbO: \((207.2 + 16.0) \ g/mol = 223.2 \ g/mol\) Theoretical yield of PbO: \(0.836 \times 223.2 = 186.5 \ g\)

3. Calculate the percent yield

To find the percent yield for this reaction, we will compare the actual yield (170.0 g of PbO) with the theoretical yield (186.5 g of PbO). Percent yield: \(\frac{Actual \ Yield}{Theoretical \ Yield} \times 100\%\) Percent yield: \(\frac{170.0 \ g}{186.5 \ g} \times 100\% \approx 91.1\%\) The percent yield for this reaction is approximately 91.1%.

Key concepts

Stoichiometry

When chemists talk about stoichiometry, they're referring to the quantitative relationship between reactants and products in a chemical reaction. This is key for predicting how much of each product you'll get from a given amount of reactants. In the example of roasting lead(II) sulfide (PbS) to get lead(II) oxide (PbO), stoichiometry helps us determine exactly how much PbO is formed from a set amount of PbS.

Here's the balanced chemical equation:
\(2 \mathrm{PbS}(s) + 3 \mathrm{O_{2}}(g) \rightarrow 2 \mathrm{PbO}(s) + 2 \mathrm{SO_{2}}(g)\).

From this, you can see that two moles of PbS react with three moles of \(\mathrm{O_{2}}\) to produce two moles of PbO alongside two moles of \(\mathrm{SO_{2}}\). Using these ratios, you can predict the amounts of reactants needed or products formed. Suppose you start with 200 grams of PbS. By using its molar mass (239.3 g/mol), you convert this mass into moles. Then, using the balanced equation, you find that the same number of moles of PbO will be produced. This predictable ratio is what makes stoichiometry so useful.

Theoretical Yield

The theoretical yield is the amount of product expected from a reaction if it proceeds perfectly and completely. It's the maximum possible amount of product (in this case, PbO) that could be formed from the given amounts of reactants under ideal conditions.

Based on the stoichiometry of the balanced equation, we already know that 0.836 moles of PbS will yield 0.836 moles of PbO under perfect circumstances. Using the molar mass of PbO (223.2 g/mol), we convert these moles into grams to find the theoretical yield:
\[0.836 \times 223.2 = 186.5 \, \text{g of PbO}\]

It's important to remember that the theoretical yield assumes complete conversion of reactants into products without any loss or side reactions, which is rarely achieved in practical scenarios.

Percent Yield

The percent yield reveals how efficient an actual reaction was compared to its theoretical potential. It's a way of quantifying any losses or inefficiencies in your procedure. To find it, you compare the actual yield (the amount of PbO actually produced) to the theoretical yield (the amount expected under perfect conditions).

In our example, the actual yield of PbO is 170.0 g. The percent yield can be calculated using the formula:

\[\text{Percent yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]
\[\text{Percent yield} = \frac{170.0}{186.5} \times 100\% \approx 91.1\%\]

This means that the reaction yielded 91.1% of the expected amount, indicating some loss likely due to side reactions, measurement inaccuracies, or incomplete reactions.