Question

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

The balanced chemical equation for the reaction between aluminum sulfide and water is: Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S. From 14.2 g of aluminum sulfide, 14.72 g of aluminum hydroxide are obtained.

Step-by-step solution

Write the unbalanced chemical equation

Aluminum sulfide (Al2S3) reacts with water (H2O) to form aluminum hydroxide (Al(OH)3) and hydrogen sulfide (H2S). We can write the chemical equation as: Al2S3 + H2O -> Al(OH)3 + H2S

Balance the chemical equation

To balance the equation, we will adjust the coefficients of the reactants and products so that the number of each type of atom is the same on both sides of the equation: Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S Now, the chemical equation is balanced.

Determine the molar masses of the compounds

In order to calculate the mass of aluminum hydroxide produced, we need the molar masses of aluminum sulfide (Al2S3) and aluminum hydroxide (Al(OH)3). Al2S3: (2 x 26.98 g/mol Al) + (3 x 32.06 g/mol S) = 53.96 g/mol Al + 96.18 g/mol S = 150.14 g/mol Al2S3 Al(OH)3: (1 x 26.98 g/mol Al) + (3 x 16 g/mol O) + (9 x 1.01 g/mol H) = 26.98 g/mol Al + 48 g/mol O + 9.09 g/mol H = 78.00 g/mol Al(OH)3

Calculate the mass of aluminum hydroxide produced

We are given 14.2 g of aluminum sulfide and need to determine the mass of aluminum hydroxide produced. We can use stoichiometry to convert the mass of aluminum sulfide to the mass of aluminum hydroxide: 1) Convert mass of aluminum sulfide (Al2S3) to moles: 14.2 g Al2S3 x (1 mol Al2S3 / 150.14 g Al2S3) = 0.0945 mol Al2S3 2) Use the balanced chemical equation to find the moles of aluminum hydroxide (Al(OH)3) produced: 0.0945 mol Al2S3 x (2 mol Al(OH)3 / 1 mol Al2S3) = 0.189 mol Al(OH)3 3) Convert moles of aluminum hydroxide (Al(OH)3) to grams: 0.189 mol Al(OH)3 x (78.00 g Al(OH)3 / 1 mol Al(OH)3) = 14.72 g Al(OH)3 So, 14.72 g of aluminum hydroxide are obtained from 14.2 g of aluminum sulfide.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, pivotal for understanding how different substances react with one another. The law of conservation of mass dictates that the mass of the reactants must be equal to the mass of the products in a chemical reaction. Consequently, every chemical equation must reflect this balance.

To achieve a balanced equation, one must ensure that the number of atoms for each element is the same on both sides of the reaction. For example, aluminum sulfide reacting with water to form aluminum hydroxide and hydrogen sulfide is represented as \(Al_2S_3 + 6H_2O \rightarrow 2Al(OH)_3 + 3H_2S\).

In this balanced equation, we can count two aluminum atoms, three sulfur atoms, and six oxygen atoms on each side, maintaining the balance required by the law of conservation of mass. Learning to balance equations involves practice and sometimes a bit of trial and error, but it's crucial for predicting the outcome of chemical reactions and performing stoichiometric calculations.

Molar Mass Calculation

The molar mass of a compound is the weight of one mole of that substance, which is the molecular weight summed up in grams. It's a bridge between the macroscopic world of grams and the microscopic world of molecules. To calculate molar mass, you add up the atomic mass of each atom in the compound, as found on the periodic table, considering the number of atoms of each element present in a molecule.

For instance, the molar mass of aluminum sulfide (\(Al_2S_3\)) is found by adding the masses of two aluminum atoms and three sulfur atoms, amounting to \(150.14\,\text{g/mol}\). Similarly, aluminum hydroxide (\(Al(OH)_3\)) has a molar mass of \(78.00\,\text{g/mol}\). It’s important to get comfortable with these calculations because they are used to convert between the amount of substance in moles and the mass in grams, rendering them indispensable in stoichiometric calculations.

Stoichiometric Calculations

Stoichiometric calculations are at the core of chemical equations, facilitating the quantification of reactants and products. These calculations start with a balanced equation, allowing chemists to understand the relationship between the amount of reactants used and the quantity of products formed.

Using the balanced reaction \(Al_2S_3 + 6H_2O \rightarrow 2Al(OH)_3 + 3H_2S\), we can deduce the stoichiometry of the reactants and products. If you begin with 14.2 grams of aluminum sulfide, stoichiometry allows you to convert this mass to moles, apply the mole ratio from the balanced equation to find the moles of aluminum hydroxide produced, and finally convert this result back to grams. The step-by-step process, starting from grams of reactants to grams of products, provides the quantitative relationship essential for comprehending the outcome of chemical reactions. Mastering stoichiometric calculations enables one to predict how much of each substance is needed or produced in a given chemical reaction, making it a cornerstone of chemistry problem-solving.