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Chapter 5: Problem 8

Aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), is used in some antiperspirants. a. How many moles of sulfur are present in \(3.0\) moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\) b. How many moles of aluminum ions are present in \(0.40\) mole of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\) c. How many moles of sulfate ions \(\left(\mathrm{SO}_{4}{ }^{2-}\right)\) are present in \(1.5\) moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) ?

Short Answer

Expert verified
a. 9.0 moles of S; b. 0.80 moles of \(\text{Al}^{3+}\); c. 4.5 moles of \(\text{SO}_4^{2-}\).

Step by step solution

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01

Title - Identify the number of sulfur atoms per formula unit for part (a)

In one formula unit of \(\text{Al}_2(\text{SO}_4)_3\), there are three sulfate ions (\(\text{SO}_4^{2-}\)). Each sulfate ion contains one sulfur atom. Therefore, there are 3 sulfur atoms per formula unit of \(\text{Al}_2(\text{SO}_4)_3\).
02

Title - Calculate the moles of sulfur for part (a)

Multiply the number of moles of \(\text{Al}_2(\text{SO}_4)_3\) by the number of sulfur atoms per formula unit: \[ \text{moles of S} = 3.0 \text{ moles of } \text{Al}_2(\text{SO}_4)_3 \times 3 \text{ sulfur atoms per formula unit} = 9.0 \text{ moles of S} \]
03

Title - Identify the number of aluminum ions per formula unit for part (b)

In one formula unit of \(\text{Al}_2(\text{SO}_4)_3\), there are two aluminum ions (\(\text{Al}^{3+}\)).
04

Title - Calculate the moles of aluminum ions for part (b)

Multiply the number of moles of \(\text{Al}_2(\text{SO}_4)_3\) by the number of aluminum ions per formula unit: \[ \text{moles of Al}^{3+} = 0.40 \text{ moles of } \text{Al}_2(\text{SO}_4)_3 \times 2 \text{ aluminum ions per formula unit} = 0.80 \text{ moles of } \text{Al}^{3+} \]
05

Title - Identify the number of sulfate ions per formula unit for part (c)

In one formula unit of \(\text{Al}_2(\text{SO}_4)_3\), there are three sulfate ions (\(\text{SO}_4^{2-}\)).
06

Title - Calculate the moles of sulfate ions for part (c)

Multiply the number of moles of \(\text{Al}_2(\text{SO}_4)_3\) by the number of sulfate ions per formula unit: \[ \text{moles of } \text{SO}_4^{2-} = 1.5 \text{ moles of } \text{Al}_2(\text{SO}_4)_3 \times 3 \text{ sulfate ions per formula unit} = 4.5 \text{ moles of } \text{SO}_4^{2-} \]

Key Concepts

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

aluminum sulfate
Aluminum sulfate, with the chemical formula \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), is an important compound used in various applications, including water purification and as an ingredient in some antiperspirants. The compound consists of aluminum ions, sulfate ions, and a complex structure that determines its properties. Each unit of aluminum sulfate contains two aluminum ions and three sulfate ions. This stoichiometric relationship allows us to perform calculations to determine the number of atoms or ions in a given amount of the compound.
sulfur atoms per compound
Understanding the number of sulfur atoms in a compound formula is crucial for mole calculations. In aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), we breakdown the structure to identify the individual components. Each sulfate ion \(\mathrm{SO}_{4}^{2-}\) contains one sulfur atom. Since there are three sulfate ions in each formula unit of aluminum sulfate, this means there are three sulfur atoms per formula unit. Therefore, if you have 3.0 moles of aluminum sulfate, you will calculate the sulfur atoms as follows: \[ 3.0 \text{ moles of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \times 3 \text{ sulfur atoms per formula unit} = 9.0 \text{ moles of sulfur} \]
aluminum ions
Aluminum sulfate contains aluminum ions, which are denoted as \(\mathrm{Al}^{3+}\). For each formula unit of aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), there are two aluminum ions. To find the moles of aluminum ions in a given amount of aluminum sulfate, use the stoichiometric ratio. For instance, in 0.40 moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), the calculation would be: \[ 0.40 \text{ moles of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \times 2 \text{ aluminum ions per formula unit} = 0.80 \text{ moles of } \mathrm{Al}^{3+} \]
sulfate ions
Sulfate ions, represented as \(\mathrm{SO}_{4}^{2-}\), play a key role in the structure of aluminum sulfate. In each formula unit of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), there are three sulfate ions. To determine the moles of sulfate ions in a specific amount of aluminum sulfate, multiply the number of moles of the compound by three. For example, in 1.5 moles of aluminum sulfate, the calculation is as follows: \[ 1.5 \text{ moles of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \times 3 \text{ sulfate ions per formula unit} = 4.5 \text{ moles of } \mathrm{SO}_{4}^{2-} \]

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Most popular questions from this chapter

Predict the products that would result from each of the following reactions and balance: a. combination: \(\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \longrightarrow\) b. decomposition: \(\mathrm{HBr}(g) \stackrel{\Delta}{\longrightarrow}\) c. single replacement: \(\mathrm{Mg}(s)+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) d. double replacement: \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) e. combustion: \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow}\)

a. Why is the following reaction called a combination reaction? $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) $$ b. Why is the following reaction called a double replacement reaction? $$ \mathrm{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q) $$

In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced: a. \(2 \mathrm{Li}(s)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{LiF}(s)\) b. \(\mathrm{Cl}_{2}(g)+2 \mathrm{KI}(a q) \longrightarrow 2 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\) c. \(2 \mathrm{Al}(s)+3 \mathrm{Sn}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \operatorname{Sn}(s)\) d. \(\mathrm{Fe}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{Cu}(s)\)

When the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor. a. Write the balanced equation for the reaction. b. How many grams of oxygen are needed to react with \(2.50 \mathrm{~g}\) of dihydrogen sulfide? c. How many grams of sulfur dioxide can be produced when \(38.5 \mathrm{~g}\) of oxygen reacts? d. How many grams of oxygen are needed to produce \(55.8 \mathrm{~g}\) of water vapor?

Determine whether each of the following equations is balanced or not balanced: a. \(\mathrm{PCl}_{3}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) b. \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. \(\mathrm{Mg}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}(s)\)
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