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Chapter 9: Problem 53

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 Icad if \(50.0 \mathrm{~kg}\) of lead oxide is heated with \(50.0 \mathrm{~kg}\) of carbon.

Short Answer

Expert verified
The expected yield of lead (Pb) when 50.0 kg of lead oxide (PbO) is heated with 50.0 kg of carbon (C) is 46.35 kg.

Step by step solution

01

Write down the balanced chemical equation

The balanced equation is given in the exercise: $$ \mathrm{PbO}(s) + \mathrm{C}(s) \rightarrow \mathrm{Pb}(l) + \mathrm{CO}(g) $$
02

Convert the given mass of reactants into moles

To find the moles of each reactant, we will use their molar masses. The molar mass of PbO is 207.2 g/mol (for Pb) + 16.0 g/mol (for O) = 223.2 g/mol. The molar mass of C is 12.01 g/mol. For PbO: $$ \text{moles of PbO} = \frac{\text{mass of PbO}}{\text{molar mass of PbO}} = \frac{50,000\text{ g}}{223.2\text{ g/mol}} = 223.97\text{ mol} $$ For C: $$ \text{moles of C} = \frac{\text{mass of C}}{\text{molar mass of C}} = \frac{50,000\text{ g}}{12.01\text{ g/mol}} = 4163.20\text{ mol} $$
03

Determine the limiting reactant

According to the balanced equation, 1 mole of PbO reacts with 1 mole of C to produce 1 mole of Pb. Now, let's compare the mole ratio of reactants available, and determine the limiting reactant. We divide the moles of each reactant by their respective stoichiometric coefficients: For PbO: $$ \frac{\text{moles of PbO}}{\text{coefficient of PbO}} = \frac{223.97\text{ mol}}{1} = 223.97 $$ For C: $$ \frac{\text{moles of C}}{\text{coefficient of C}} = \frac{4163.20\text{ mol}}{1} = 4163.20 $$ Since 223.97 is smaller than 4163.20, PbO is the limiting reactant.
04

Calculate the expected yield of lead (Pb)

Now that we know PbO is the limiting reactant, we can calculate the expected yield of Pb using the mole ratio of PbO to Pb in the balanced equation: $$ \text{moles of Pb} = \frac{\text{moles of PbO}}{1} = \frac{223.97\text{ mol}}{1} = 223.97\text{ mol} $$ Next, we will convert the moles of Pb into mass using the molar mass of Pb (207.2 g/mol): $$ \text{mass of Pb} = \text{moles of Pb} \times \text{molar mass of Pb} = 223.97\text{ mol} \times 207.2\text{ g/mol} = 46350.96\text{ g} $$ Converting the value into kilograms, we have: $$ \text{yield of Pb} = \frac{46350.96\text{ g}}{1000\text{ g/kg}} = 46.35\text{ kg} $$ So, the expected yield of lead is 46.35 kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equations
Chemical equations are symbolic representations of a chemical reaction. They include reactants on the left and products on the right, separated by an arrow indicating the direction of the reaction.
In our example, the equation is: PbO(s) + C(s) 鈫� Pb(l) + CO(g).
This tells us that lead(II) oxide (PbO) reacts with carbon (C) to produce lead (Pb) and carbon monoxide (CO).
- **Reactants**: These are substances consumed in a reaction. Here, PbO and C are the reactants. - **Products**: Substances formed as a result of the reaction. Pb and CO are the products here. A balanced chemical equation ensures that the number of each type of atom is the same on both sides of the equation. This is crucial for stoichiometric calculations, as it maintains the law of conservation of mass.
Limiting Reactant
The limiting reactant in a chemical reaction is the substance that is completely used up first, stopping the reaction from continuing, as it limits the amount of products formed. In the given exercise, we determine the limiting reactant by comparing the moles of each reactant available with their stoichiometric coefficients from the balanced equation.
Here's how it works: - Calculate moles of each reactant using their masses and molar masses. - Compare these mole ratios with the stoichiometric coefficients in the balanced equation. - The reactant with the smallest ratio is the limiting reactant. In our task, PbO limits the reaction. We have fewer moles of PbO compared to carbon, making it the limiting reactant and dictating that 223.97 moles of Pb can be produced. Identifying the limiting reactant is crucial in predicting the maximum product yield of any reaction.
Molar Mass
Molar mass is the mass of a given substance (element or compound) divided by the amount of substance (mol). It is essential in converting between grams and moles, allowing us to perform stoichiometric calculations effectively.
Consider these points: - **Calculation**: For a compound, add the molar masses of its elements. For PbO, the molar mass is computed as the sum of lead (207.2 g/mol) and oxygen (16.0 g/mol), resulting in 223.2 g/mol. - **Conversion**: Use molar masses to convert between mass and moles. For example, to find moles of PbO from a mass of 50.0 kg, convert kg to grams, then divide by molar mass. - **Practical Application**: It's essential in calculating amounts in reactions, like predicting product yields. Understanding molar mass is fundamental in stoichiometry, as it bridges the gap between the macro world (measured in grams) and the micro world (measured in moles). This bridge is critical for accurate chemical reaction predictions.

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Most popular questions from this chapter

For the balanced chemical equation for the combination reaction of hydrogen gas and oxygen gas $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ explain why we know that \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) reacting with \(1 \mathrm{~g}\) of \(\mathrm{O}_{2}\) will not result in the production of \(2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\)

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), cmits a large quantity of energy when it reacts with oxygen, which has led to hydrazine's use as a fuel for rockets: $$ \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ How many moles of each of the gaseous products are produced when \(20.0 \mathrm{~g}\) of pure hydrazine is ignited in the presence of \(20.0 \mathrm{~g}\) of pure oxygen? How many grams of each product are produced?

A \(0.4230-g\) sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to \(0.2339 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

For each of the following balanced chemical equations, calculate how many grams of the product(s) would be produced by complete reaction of 0.125 mole of the first reactant. a. \(\operatorname{AgNO}_{3}(a q)+\mathrm{LiOH}(a q) \rightarrow \mathrm{AgOH}(s)+\mathrm{LiNO}_{3}(a q)\) b. \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{CaCl}_{2}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{CaSO}_{4}(s)\) c. \(\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\)

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a 20.4 -gram sample of pentane is burned in excess oxygen, what mass of water can be produced, assuming \(100 \%\) yield?
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