Problem 174 - Introductory Chemistry Atoms First

Question

Consider the unbalanced chemical equation \(\mathrm{Al}_{2} \mathrm{~S}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{~S}\) If \(56.0 \mathrm{~g}\) of aluminum sulfide reacts with \(48.2 \mathrm{~g}\) of water, (a) Which is the excess reactant? (b) What mass in grams of the excess reactant remains after the reaction is complete?

Step-by-Step Solution

Answer

(a) The excess reactant is H2O (Water). (b) The mass of the excess reactant remaining after the reaction is complete is 7.96 g of H2O.

Step 1: Balanced Chemical Equation

To balance the chemical equation, we need each element to have an equal number of atoms on both sides of the equation: \(\mathrm{Al}_{2} \mathrm{S}_{3} + 6\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{Al}(\mathrm{OH})_{3} + 3\mathrm{H}_{2} \mathrm{S}\) Now that we have the balanced equation, we can proceed with determining the limiting and excess reactants.

Step 2: Calculating Mol

Determine the mols of each reactant by dividing the given mass by their molar mass. For Aluminum sulfide (Al2S3): Molar mass of Al2S3 = \(2 \times 26.98\, \text{g/mol (Al)} + 3 \times 32.07\, \text{g/mol (S)} = 150.16\, \text{g/mol}\) Mols of Al2S3 = \(\frac{56.0\, \text{g}}{150.16\, \text{g/mol}} = 0.373\, \text{mol}\) For Water (H2O): Molar mass of H2O = \(2 \times 1.01\, \text{g/mol (H)} + 1 \times 16.00\, \text{g/mol (O)} = 18.02\, \text{g/mol}\) Mols of H2O = \(\frac{48.2\, \text{g}}{18.02\, \text{g/mol}} = 2.68\, \text{mol}\)

Step 3: Identifying Limiting and Excess Reactants

Now, we will determine the molar ratios of the two reactants and compare them to the stoichiometric coefficients in the balanced equation. Molar ratio of Al2S3 to H2O = \(\frac{0.373\, \text{mol (Al2S3)}}{2.68\, \text{mol (H2O)}} = 0.139\) Stoichiometric ratio (from the balanced equation) = \(\frac{1\, \text{mol (Al2S3)}}{6\, \text{mol (H2O)}} = 0.167\) Since the molar ratio (0.139) is less than the stoichiometric ratio (0.167), Al2S3 is the limiting reactant and H2O is the excess reactant.

Step 4: Calculate Mass of Excess Reactant Remaining

First, let's determine how many moles of H2O were used by the reaction: Moles of H2O used = \(0.373\, \text{mol (Al2S3)} \times \frac{6\, \text{mol (H2O)}}{1\, \text{mol (Al2S3)}} = 2.238\, \text{mol (H2O)}\) Now, subtract the moles of H2O used from the initial amount to find moles of H2O remaining: Moles of H2O remaining = \(2.68\, \text{mol (H2O)} - 2.238\, \text{mol (H2O)} = 0.442\, \text{mol (H2O)}\) Finally, convert the moles of H2O remaining back to grams using the molar mass: Mass of H2O remaining = \(0.442\, \text{mol} \times 18.02\, \text{g/mol} = 7.96\, \text{g}\) (a) The excess reactant is H2O (Water). (b) The mass of the excess reactant remaining after the reaction is complete is 7.96 g of H2O.

Key Concepts

Understanding Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It's based on the conservation of mass and the concept that atoms are neither created nor destroyed in chemical reactions. Students often encounter stoichiometry when they must calculate how much of a product is formed from given amounts of reactants or when determining the limiting and excess reactants in a chemical equation.

When working through stoichiometric problems, we use a balanced chemical equation as our 'recipe' for the reaction. This equation tells us in what ratios the molecules react and are produced. In the provided exercise, stoichiometry is the tool that helps us determine the excess reactant and its remaining mass after the reaction.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, as it ensures that the Law of Conservation of Mass is upheld. It involves making sure that there are equal numbers of each type of atom on both sides of the reaction arrow. A balanced chemical equation is essential for stoichiometric calculations because it gives the exact ratio in which reactants combine to form products.

The given textbook solution demonstrates how to convert an unbalanced equation to a balanced one by adjusting the coefficients of the reactants and products. Remember, we can only change the coefficients in front of the compounds to balance equations, not the subscripts within the compounds. This process may require a bit of trial and error, but it's important for understanding the quantitative aspects of chemical reactions.

Molar Mass Calculations

Molar mass is defined as the mass of one mole of a substance. It's an essential concept for quantifying reactions in chemistry and is expressed in grams per mole (g/mol). In the context of limiting and excess reactants, we use molar mass to convert the mass of reactants to moles, allowing us to use the balanced equation to find the stoichiometry of the reaction.

As illustrated in the solution, calculating the molar mass of a compound involves adding together the atomic masses (from the periodic table) of all atoms present in the molecule, thereby allowing for a step-by-step analysis of the reaction. Without accurate molar mass calculations, it would be impossible to perform stoichiometric computations, and we could not accurately determine the limiting reactant or the excess reactant in a given reaction.