Problem 72 - Chemistry: The Central Science

Question

Aluminum hydroxide reacts with sulfuric acid as follows: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(0.500 \mathrm{~mol}\) \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

Step-by-Step Solution

Answer

The limiting reactant is \(\mathrm{H_2SO_4}\), under these conditions \(0.167\ \mathrm{mol}\) of \(\mathrm{Al_2(SO_4)_3}\) can form, and \(0.250\ \mathrm{mol}\) of the excess reactant \(\mathrm{Al(OH)_3}\) remain after the completion of the reaction.

Step 1: Identify the balanced chemical equation

The balanced chemical equation for the reaction is: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$

Step 2: Calculate the mole ratio

First, find the mole ratio of \(\mathrm{Al(OH)_3}\) to \(\mathrm{H_2SO_4}\) required for the reaction to proceed: $$ \frac{\text{moles of } \mathrm{Al}(\mathrm{OH})_{3}}{\text{moles of } \mathrm{H}_{2} \mathrm{SO}_{4}} = \frac{2}{3} $$

Step 3: Determine the limiting reactant

To determine the limiting reactant, compare the given amounts of reactants and their mole ratio. A ratio less than the required mole ratio will indicate the limiting reactant: $$ \frac{0.500\ \mathrm{mol}\ \mathrm{Al}(\mathrm{OH})_{3}}{0.500\ \mathrm{mol}\ \mathrm{H}_{2} \mathrm{SO}_{4}} = 1 $$ Since the given mole ratio (1) is greater than the required mole ratio (\(\frac{2}{3}\)), \(\mathrm{H_2SO_4}\) is the limiting reactant.

Step 4: Calculate the moles of \(\mathrm{Al_2(SO_4)_3}\) formed

Now, we can find the moles of \(\mathrm{Al_2(SO_4)_3}\) formed using the stoichiometry from the balanced equation and the moles of limiting reactant (\(\mathrm{H_2SO_4}\)): $$ \text{moles of } \mathrm{Al_2(SO_4)_3} = \frac{\text{moles of } \mathrm{H_2SO_4}}{3} \times 1 = \frac{0.500\ \mathrm{mol}}{3} = 0.167\ \mathrm{mol} $$

Step 5: Calculate the moles of excess reactant remaining

To find the moles of excess reactant (\(\mathrm{Al(OH)_3}\)) remaining, we need to find the moles of \(\mathrm{Al(OH)_3}\) that reacted and subtract it from the initial given moles: $$ \text{moles of } \mathrm{Al}(\mathrm{OH})_3\text{ that reacted} = \frac{1}{2} \times \text{moles of } \mathrm{H_2SO_4} = \frac{1}{2} \times 0.500\ \mathrm{mol} = 0.250\ \mathrm{mol} $$ Hence, the moles of excess reactant remaining: $$ \text{moles of remaining } \mathrm{Al}(\mathrm{OH})_{3} = \text{initial moles of }\ \mathrm{Al}(\mathrm{OH})_{3} - \text{moles of } \mathrm{Al}(\mathrm{OH})_{3}\text{ that reacted} = 0.500\ \mathrm{mol} - 0.250\ \mathrm{mol} = 0.250\ \mathrm{mol} $$ In conclusion, the limiting reactant is \(\mathrm{H_2SO_4}\), under these conditions \(0.167\ \mathrm{mol}\) of \(\mathrm{Al_2(SO_4)_3}\) can form, and \(0.250\ \mathrm{mol}\) of the excess reactant \(\mathrm{Al(OH)_3}\) remain after the completion of the reaction.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that allows us to understand the quantitative relationships in a chemical reaction. It provides a systematic way to calculate the amounts of reactants and products involved. When performing stoichiometric calculations, we rely on the balanced chemical equation, which gives us the mole ratios of the different chemicals involved.
This process involves three primary steps: writing down the balanced chemical equation, determining the mole ratio from this equation, and using these ratios to solve for unknown quantities.

• First, ensure that your chemical equation is balanced; all the elements on the reactant side are accounted for in the product side.
• Second, use the balanced equation to determine the mole ratio between the reactants and the products.
• Finally, use these ratios to calculate any unknown quantity, such as the amount of product formed or the excess reactants remaining.

By understanding and applying stoichiometry, you can predict how much product will form and identify limiting and excess reactants in chemical reactions.

Mole Ratio

The concept of a mole ratio is crucial when interpreting chemical equations. It's derived directly from a balanced chemical equation and indicates the proportional relationship between the reactants and products involved in a reaction.
Given the balanced equation for the reaction: \[ 2 \mathrm{Al} (\mathrm{OH})_{3} + 3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2} \left(\mathrm{SO}_{4}\right)_{3} + 6 \mathrm{H}_{2} \mathrm{O} \]
The mole ratio tells us that every 2 moles of aluminum hydroxide reacts with 3 moles of sulfuric acid. This ratio is vital for determining which reactant will be used up first (the limiting reactant) and how much of the product can be expected.

• To apply this, compare the initial amounts of your reactants and set up a proportion using their mole ratio.
• If the proportion of the actual amounts is less than the ratio, the reactant on the numerator is the limiting reactant.
• Using these mole ratios is key to making predictions about when the reaction will stop, what is leftover, and how much product is formed.

Understanding mole ratios helps in converting between moles and masses of compounds, a vital part of stoichiometry.

Balanced Chemical Equation

A balanced chemical equation is the starting point for any chemical reaction analysis. It clearly represents the conservation of mass, where the number of atoms of each element is the same on both sides of the equation. This essential principle allows chemists to predict the amounts of products formed in a reaction given a specific amount of reactants and is crucial for stoichiometric calculations.
In the given exercise, the balanced equation is: \[ 2 \mathrm{Al} (\mathrm{OH})_{3} + 3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2} \left(\mathrm{SO}_{4}\right)_{3} + 6 \mathrm{H}_{2} \mathrm{O} \]
Here, the coefficients (2, 3, 1, and 6) ensure that the equation is balanced, representing the exact stoichiometry of the reaction.

• Each coefficient corresponds to the number of moles needed or produced.
• Balancing an equation involves changing these coefficients, not the chemical formula of the compounds, to ensure atom conservation.
• The balanced equations provide the framework for calculating how much reactant is needed to produce a desired amount of product.

Ensuring equations are balanced is a fundamental step that must be carried out before performing any further stoichiometric calculations or solving for limiting reactants.