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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 is: Al2S3 (s) + 6H2O (l) -> 2Al(OH)3 (aq) + 3H2S (g) (b) Approximately 14.77 g of aluminum hydroxide are obtained from 14.2 g of aluminum sulfide.

Step by step solution

01

Write the balanced chemical equation

The unbalanced chemical equation for the reaction can be written as: Al2S3 (s) + H2O (l) -> Al(OH)3 (aq) + H2S (g) To balance the equation, we need to make sure there are the same number of each element on both the reactant side and the product side. After balancing, the equation becomes: Al2S3 (s) + 6H2O (l) -> 2Al(OH)3 (aq) + 3H2S (g)
02

Calculate the moles of aluminum sulfide

To calculate the moles of aluminum sulfide, we can use the given mass (14.2 g) and molar mass of aluminum sulfide, which is approximately 150 g/mol (26.98 g/mol for aluminum and 32.07 g/mol for sulfur). Moles of Al2S3 = (14.2 g) / (150 g/mol) = 0.0947 mol
03

Use the mole ratio to find the moles of aluminum hydroxide produced

From the balanced equation, we can see that 1 mole of Al2S3 produces 2 moles of Al(OH)3. So, with the available moles of aluminum sulfide: Moles of Al(OH)3 = Moles of Al2S3 脳 (2 moles of Al(OH)3 / 1 mole of Al2S3) = 0.0947 mol 脳 2 = 0.1894 mol
04

Calculate the mass of aluminum hydroxide produced

To find the mass of aluminum hydroxide produced, we can use the moles calculated in the previous step and the molar mass of aluminum hydroxide. The molar mass of Al(OH)3 is approximately 78 g/mol (26.98 g/mol for aluminum and 3 脳 16 g/mol for oxygen plus 3 脳 1 g/mol for hydrogen): Mass of Al(OH)3 = Moles of Al(OH)3 脳 Molar mass of Al(OH)3 Mass of Al(OH)3 = (0.1894 mol) 脳 (78 g/mol) = 14.77 g So, from 14.2 g of aluminum sulfide, approximately 14.77 g of aluminum hydroxide are produced.

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Most popular questions from this chapter

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As \(\mathrm{NO}\) is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called " \(\mathrm{NO}_{\mathrm{x}}^{\prime \prime}\) gases. In \(2007,\) the United States emitted an estimated 22 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

What parts of balanced chemical equations give information about the relative numbers of moles of reactants and products involved in a reaction?

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