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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 → 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 → 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 → 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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Most popular questions from this chapter

Valproic acid, used to treat seizures and bipolar disorder, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} .\) A \(0.165-\mathrm{g}\) sample is combusted to produce \(0.166 \mathrm{~g}\) of water and \(0.403 \mathrm{~g}\) of carbon dioxide. What is the empirical formula for valproic acid? If the molar mass is \(144 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula?

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