Question

Sulfuric acid reacts with aluminum hydroxide by double replacement. a. If 30.0 g of sulfuric acid react with 25.0 g of aluminum hydroxide, identify the limiting reactant. b. Determine the mass of excess reactant remaining. c. Determine the mass of each product formed. Assume 100% yield.

Short answer

a. H₂SO₄ is the limiting reactant. b. 17.1 g of Al(OH)₃ remains. c. 34.9 g of Al₂(SO₄)₃ and 11.0 g of H₂O are produced.

Step-by-step solution

Write and balance the chemical equation

First, determine the reactants and products. The chemical reaction for the given problem is between sulfuric acid (H₂SO₄) and aluminum hydroxide (Al(OH)₃). The balanced chemical equation is: Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O Al(OH)₃ (aq) + 3 H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 6 H₂O (l)

Calculate moles of each reactant

Calculate the moles of each reactant using their molar masses. Molar mass of H₂SO₄ = 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol Moles of H₂SO₄ = 30.0g / 98.09 g/mol = 0.306 moles Molar mass of Al(OH)₃ = 26.98 + 3(16.00 + 1.01) = 78.00 g/mol Moles of Al(OH)₃ = 25.0g / 78.00 g/mol = 0.321 moles

Identify the limiting reactant

Using the balanced chemical equation, determine the reactant that will be completely consumed first (the limiting reactant). Based on stoichiometry: 1 mole of Al(OH)₃ reacts with 3 moles of H₂SO₄ Therefore, 0.321 moles of Al(OH)₃ require 3 * 0.321 = 0.963 moles of H₂SO₄. Since we only have 0.306 moles of H₂SO₄ available, H₂SO₄ is the limiting reactant.

Calculate the mass of the excess reactant remaining

Find out how much Al(OH)₃ is left unreacted using the limiting reactant (H₂SO₄). Moles of Al(OH)₃ used = 0.306 moles H₂SO₄ * (1 mole Al(OH)₃ / 3 moles H₂SO₄) = 0.102 moles Moles of excess Al(OH)₃ = 0.321 moles - 0.102 moles = 0.219 moles Mass of excess Al(OH)₃ = 0.219 moles * 78.00 g/mol = 17.1 g

Calculate the mass of each product formed

Determine the mass of Al₂(SO₄)₃ and H₂O produced using the stoichiometry of the reaction. Moles of Al₂(SO₄)₃ formed from limiting reactant H₂SO₄: Moles of Al₂(SO₄)₃ = 0.306 moles H₂SO₄ * (1 mole Al₂(SO₄)₃ / 3 moles H₂SO₄) = 0.102 moles Mass of Al₂(SO₄)₃ = 0.102 moles * 342.15 g/mol = 34.9 g Moles of H₂O formed from limiting reactant H₂SO₄: Moles of H₂O = 0.306 moles H₂SO₄ * (6 moles H₂O / 3 moles H₂SO₄) = 0.612 moles Mass of H₂O = 0.612 moles * 18.02 g/mol = 11.0 g

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that gets completely used up first. This reactant limits the amount of product that can be formed. To identify the limiting reactant, we compare the mole ratio of the reactants used in the balanced chemical equation. For example, in the reaction between sulfuric acid (H₂SO₄) and aluminum hydroxide (Al(OH)₃), we discovered that H₂SO₄ is the limiting reactant.
By calculating the moles of each reactant and comparing their ratios, we saw that the given amount of H₂SO₄ wasn't enough to react with all the Al(OH)₃ available, thus making H₂SO₄ the limiting reactant.

• Key point: The reactant that is entirely consumed first is the limiting reactant.

Balanced Chemical Equation

A balanced chemical equation ensures that the number of atoms of each element is the same on both sides of the reaction. This is crucial for accurately solving stoichiometry problems. For our reaction:
Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
It's balanced because it has equal numbers of each type of atom on both sides.

• Balancing equations involves adjusting coefficients to align atom counts.
• Double-check atom counts for precision.

Molar Mass

Molar mass is the mass of one mole of a substance and is essential for converting between grams and moles. The molar masses for relevant substances are:

• H₂SO₄: 98.09 g/mol
• Al(OH)₃: 78.00 g/mol
• Al₂(SO₄)₃: 342.15 g/mol
• H₂O: 18.02 g/mol

By dividing mass by molar mass, we convert grams to moles, which is necessary for stoichiometric calculations.

Excess Reactant

The excess reactant is the substance that remains after the limiting reactant is completely consumed. In our problem, Al(OH)₃ is the excess reactant. By using the moles of the limiting reactant (H₂SO₄) and the reaction's stoichiometry, we calculated the moles of Al(OH)₃ that actually reacted. We then subtracted this from the initial moles of Al(OH)₃ to find the excess.

• Key point: Excess reactants are not fully used up in the reaction.

Mass of Products

To find the mass of products formed, we use the moles of the limiting reactant and the stoichiometry of the balanced equation. In our case, H₂SO₄ determined the amounts of Al₂(SO₄)₃ and H₂O formed. Using stoichiometric coefficients and molar masses, we calculated:

• Mass of Al₂(SO₄)₃ formed: 34.9 g
• Mass of H₂O formed: 11.0 g

Ensure mass calculations are based on the limiting reactant to obtain accurate results.