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Chapter 3: Problem 65

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from \(14.2 \mathrm{~g}\) of aluminum sulfide?

Short Answer

Expert verified
(a) The balanced chemical equation for the reaction between aluminum sulfide and water is: Al2S3 + 6H2O 鈫� 2Al(OH)3 + 3H2S (b) Approximately 14.76 grams of aluminum hydroxide are obtained when 14.2 grams of aluminum sulfide reacts with water.

Step by step solution

01

Write the Unbalanced Chemical Equation

The reaction between aluminum sulfide (Al2S3) and water (H2O) produces aluminum hydroxide (Al(OH)3) and hydrogen sulfide (H2S). The unbalanced chemical equation is: Al2S3 + H2O 鈫� Al(OH)3 + H2S
02

Balance the Chemical Equation

To balance the equation, we'll adjust the coefficients so that the number of atoms of each element is the same on both sides. Al2S3 + 6H2O 鈫� 2Al(OH)3 + 3H2S Now the equation is balanced: there are 2 aluminum, 6 oxygen, 6 hydrogen, and 3 sulfur atoms on both sides of the equation.
03

Find Moles of Aluminum Sulfide Given

To determine the number of moles of aluminum sulfide given, we'll use the mass of aluminum sulfide and its molar mass. The molar mass of Al2S3 is: (2 x 26.98 g/mol for Al) + (3 x 32.06 g/mol for S) = 150.16 g/mol We are given 14.2 grams of aluminum sulfide, so we can find the moles as follows: \( \text{moles of Al2S3} = \frac{\text{mass of Al2S3}}{\text{molar mass of Al2S3}} \) \( \text{moles of Al2S3} = \frac{14.2 \mathrm{~g}}{150.16 \mathrm{~g/mol}} \approx 0.0946 \mathrm{~mol} \)
04

Calculate Moles of Aluminum Hydroxide Formed

Using the balanced chemical equation, we can find the moles of aluminum hydroxide formed through stoichiometry. For each mole of aluminum sulfide, two moles of aluminum hydroxide are formed: \( \text{moles of Al(OH)3} = 2 \times \text{moles of Al2S3} \) \( \text{moles of Al(OH)3} = 2 \times 0.0946 \mathrm{~mol} \approx 0.1892 \mathrm{~mol} \)
05

Calculate Mass of Aluminum Hydroxide Formed

Finally, we will determine the mass of aluminum hydroxide formed using the number of moles we found in the previous step and the molar mass of Al(OH)3 which is: (1 x 26.98 g/mol for Al) + (3 x 15.999 g/mol for O) + (3 x 1.008 g/mol for H) = 78.003 g/mol \( \text{mass of Al(OH)3} = \text{moles of Al(OH)3} \times \text{molar mass of Al(OH)3} \) \( \text{mass of Al(OH)3} = 0.1892 \mathrm{~mol} \times 78.003 \mathrm{~g/mol} \approx 14.76 \mathrm{~g} \) Therefore, approximately 14.76 grams of aluminum hydroxide are obtained when 14.2 grams of aluminum sulfide reacts with water.

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Key Concepts

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

Stoichiometry
Stoichiometry is like a recipe in chemistry. It helps us understand how different chemical reactants and products relate in a reaction. In our example with aluminum sulfide reacting with water, stoichiometry tells us the proportions needed. By using a balanced chemical equation, we know exactly how much of one substance will react with another or how much product will be formed.

When dealing with stoichiometry:
  • Identify the balanced equation, which shows the relationship between reactants and products.
  • Use the mole ratios to determine quantities. For instance, from the balanced equation, each mole of aluminum sulfide produces two moles of aluminum hydroxide.
Understanding stoichiometry allows you to predict the outcome of the reaction in terms of mass, enabling precise calculations.
Chemical Equation Balancing
Balancing a chemical equation ensures that the same number of each type of atom appears on both sides of the reaction. This is crucial because of the law of conservation of mass, which states that matter cannot be created or destroyed.

In our example, the unbalanced equation was:\[\text{Al}_2\text{S}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{H}_2\text{S}\]To balance it, we adjusted coefficients:\[\text{Al}_2\text{S}_3 + 6\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3 + 3\text{H}_2\text{S}\]This guarantees an equal number of aluminum, sulfur, hydrogen, and oxygen atoms on both sides.

Balancing equations involves trial and error, but understanding how atoms rearrange during reactions is key.
Molar Mass
Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It's essential for converting between grams and moles when solving chemistry problems.

To find molar mass, add up the atomic masses of all atoms in a compound. For instance, the molar mass of aluminum sulfide (Al鈧係鈧�) is calculated by:
  • Adding 2 times the atomic mass of aluminum (26.98 g/mol).
  • Adding 3 times the atomic mass of sulfur (32.06 g/mol).
This gives us a molar mass of 150.16 g/mol for Al鈧係鈧�.

Knowing this, you can convert given grams of a substance into moles to use in stoichiometric calculations, such as determining how many moles of a product you will get from a reaction.

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Most popular questions from this chapter

Write "true" or "false" for each statement. (a) We balance chemical equations as we do because energy must be conserved. (b) If the reaction \(2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g)\) goes to completion and all \(\mathrm{O}_{3}\) is converted to \(\mathrm{O}_{2}\), then the mass of \(\mathrm{O}_{3}\) at the beginning of the reaction must be the same as the mass of \(\mathrm{O}_{2}\) at the end of the reaction. \((\mathbf{c})\) You can balance the "water-splitting" reaction \(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) by writing it this way: \(\mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\).

Boron nitride, \(\mathrm{BN}\), is an electrical insulator with remarkable thermal and chemical stability. Its density is \(2.1 \mathrm{~g} / \mathrm{cm}^{3}\). It can be made by reacting boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3}\), with ammonia. The other product of the reaction is water. (a) Write a balanced chemical equation for the synthesis of BN. (b) If you made \(225 \mathrm{~g}\) of boric acid react with \(150 \mathrm{~g}\) ammonia, what mass of BN could you make? (c) Which reactant, if any, would be left over, and how many moles of leftover reactant would remain? (d) One application of \(\mathrm{BN}\) is as thin film for electrical insulation. If you take the mass of BN from part (a) and make a \(0.4 \mathrm{~mm}\) thin film from it, what area, in \(\mathrm{cm}^{2}\), would it cover?

Write balanced chemical equations for \((\mathbf{a})\) the complete combustion of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right),\) a common organic solvent; (b) the decomposition of solid mercury (I) carbonate into carbon dioxide gas, mercury, and solid mercury oxide; (c) the combination reaction between sulphur dioxide gas and liquid water to produce sulfurous acid.

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(\mathrm{B}_{2} \mathrm{~S}_{3}(s),\) reacts violently with water to form dissolved boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3},\) and hydrogen sulfide gas. (c) Phosphine, \(\mathrm{PH}_{3}(g)\), combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen. (e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{3} \mathrm{O}\), molar mass \(=62.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2}\), molar mass \(=32.0 \mathrm{~g} / \mathrm{mol}\)
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