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Chapter 8: Problem 23

For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume there is more than enough of the other reactant. $$ 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) $$ (a) \(2.4 \mathrm{~mol} \mathrm{PbS}\) (b) \(2.4 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(5.3 \mathrm{~mol} \mathrm{PbS}\) (d) \(5.3 \mathrm{~mol} \mathrm{O}_{2}\)

Short Answer

Expert verified
From 2.4 mol PbS, 2.4 mol PbO and 2.4 mol SO2 form. From 2.4 mol O2, 1.6 mol PbO and 1.6 mol SO2 form. From 5.3 mol PbS, 5.3 mol PbO and 5.3 mol SO2 form. From 5.3 mol O2, 3.53 mol PbO and 3.53 mol SO2 form.

Step by step solution

01

- Understanding the Stoichiometry

First, understand the stoichiometry of the balanced reaction. It shows that 2 moles of PbS react with 3 moles of O2 to produce 2 moles of PbO and 2 moles of SO2.
02

- Calculating Moles of Products from Moles of PbS

Use the stoichiometric coefficients to find the moles of products for (a) and (c). For every 2 moles of PbS, 2 moles of PbO and 2 moles of SO2 are produced.
03

- Calculating Product for (a) 2.4 mol PbS

For (a), we have 2.4 moles of PbS. The molar ratio for PbS to PbO and PbS to SO2 is 1:1. Therefore, 2.4 moles of PbS produce 2.4 moles of PbO and 2.4 moles of SO2.
04

- Calculating Moles of Products from Moles of O2

The reaction shows that for every 3 moles of O2, 2 moles of PbO and 2 moles of SO2 form. This is a 3:2 ratio.
05

- Calculating Product for (b) 2.4 mol O2

For (b), with 2.4 moles of O2, and using the molar ratio of 3:2 for O2 to PbO and SO2, we have (2.4 moles O2) * (2 moles PbO / 3 moles O2) = 1.6 moles PbO, and similarly, 1.6 moles SO2.
06

- Calculating Product for (c) 5.3 mol PbS

For (c), with 5.3 moles of PbS and a 1:1 molar ratio, there will be 5.3 moles of PbO and 5.3 moles of SO2 produced.
07

- Calculating Product for (d) 5.3 mol O2

For (d), we have 5.3 moles of O2. Applying the 3:2 molar ratio for O2 to PbO and SO2, we get (5.3 moles O2) * (2 moles PbO / 3 moles O2) = 3.53 moles PbO and similarly, 3.53 moles SO2.

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Key Concepts

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

Stoichiometric Calculations
Stoichiometric calculations are mathematical techniques used to determine the quantities of reactants and products involved in a chemical reaction. These calculations are based on the balanced chemical equation, which provides the necessary molar ratios of the substances involved.

For instance, using the provided exercise, if we have 2.4 moles of lead(II) sulfide (PbS), we understand from the balanced equation that there is a direct one-to-one molar relationship between PbS and the products lead(II) oxide (PbO) and sulfur dioxide (SO2). So, 2.4 moles of PbS will produce 2.4 moles of PbO and 2.4 moles of SO2.

To improve understanding of stoichiometric calculations, it's helpful to:
  • Write down the balanced chemical equation.
  • Determine the mole ratio from the coefficients.
  • Use unit conversions to align the calculations with the desired units.
  • Apply the mole ratio to calculate the quantities of products or reactants needed.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry deals with quantitatively evaluating the proportions of elements and compounds involved in chemical reactions. It's the core principle guiding stoichiometric calculations.

In our exercise, the reaction between PbS and O2 is a great example. The balanced chemical equation indicates that, in the presence of sufficient amounts of oxygen, PbS will react to form PbO and SO2 in a ratio defined by the reaction coefficients. This is crucial to predict how much product will form from a given amount of reactant.

When improving students' grasp on chemical reaction stoichiometry, focus on:
  • Understanding the law of conservation of mass.
  • Knowing how to balance chemical equations.
  • Recognizing the importance of molar ratios in predicting the amounts of products and reactants.
Mole Concept
The mole concept is a fundamental pillar in chemistry that provides a bridge between the atomic or molecular world and the macroscopic world we observe. One mole is defined as exactly 6.022 x 1023 representative particles (such as atoms, ions, or molecules), known as Avogadro's number.

In the context of the exercise, when we say '2.4 moles of PbS,' we're referring to 2.4 times Avogadro's number of PbS formula units. Understanding this concept allows us to perform calculations like those in the provided steps, converting moles to grams if necessary, and vice versa.

Enhance comprehension of the mole concept by:
  • Explaining Avogadro's number and its significance.
  • Connecting the concept of moles to real-world quantities through examples and practice problems.
  • Stressing the consistency of the mole as a counting unit across different substances.

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Most popular questions from this chapter

Why is reaction stoichiometry important? Cite some examples in your answer.

Explain the relationship between the sign of \(\Delta H_{\mathrm{rxn}}\) and whether a reaction is exothermic or endothermic.

For the reaction shown, calculate the theoretical yield of product in moles for each of the initial quantities of reactants. $$ 2 \mathrm{Mn}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MnO}_{3}(s) $$ (a) \(2 \mathrm{~mol} \mathrm{Mn} ; 2 \mathrm{~mol} \mathrm{O}_{2}\) (b) \(4.8 \mathrm{~mol} \mathrm{Mn} ; 8.5 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(0.114 \mathrm{~mol} \mathrm{Mn} ; 0.161 \mathrm{~mol} \mathrm{O}_{2}\) (d) \(27.5 \mathrm{~mol} \mathrm{Mn} ; 43.8 \mathrm{~mol} \mathrm{O}_{2}\)

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. $$ 4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s) $$ (a) \(1 \mathrm{~mol} \mathrm{Cr} ; 1 \mathrm{~mol} \mathrm{O}_{2}\) (b) \(4 \mathrm{~mol} \mathrm{Cr} ; 2.5 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(12 \mathrm{~mol} \mathrm{Cr} ; 10 \mathrm{~mol} \mathrm{O}_{2}\) (d) \(14.8 \mathrm{~mol} \mathrm{Cr} ; 10.3 \mathrm{~mol} \mathrm{O}_{2}\)

Sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right)\) reacts with water to form sodium hydroxide and oxygen gas. Write a balanced equation for the reaction and determine how much oxygen in grams is formed by the complete reaction of \(35.23 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}_{2}\).
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