Problem 78 - Chemistry The Central Science In Si Units

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 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 H₂SO₄, forming 0.167 mol Al₂(SO₄)₃ with 0.167 mol Al(OH)₃ remaining.

Step 1: Identify Mole Ratios from Balanced Equation

The balanced chemical equation is \(2 \mathrm{Al} (\mathrm{OH})_3 + 3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + 6 \mathrm{H}_2 \mathrm{O}\). From this equation, the mole ratio between \(\mathrm{Al} (\mathrm{OH})_3\) and \(\mathrm{H}_2 \mathrm{SO}_4\) is 2:3.

Step 2: Calculate Required Moles of Each Reactant

To use the entire amount of \(0.500 \text{ mol} \) \(\mathrm{Al} (\mathrm{OH})_3\), you need \(0.500 \text{ mol} \times \frac{3}{2} = 0.750 \text{ mol} \) of \(\mathrm{H}_2 \mathrm{SO}_4\). To fully react with the given \(0.500 \text{ mol} \) \(\mathrm{H}_2 \mathrm{SO}_4\), you need \(0.500 \text{ mol} \times \frac{2}{3} = 0.333 \text{ mol} \) of \(\mathrm{Al} (\mathrm{OH})_3\).

Step 3: Identify Limiting Reactant

Since we have only \(0.500 \text{ mol}\) of both \(\mathrm{Al} (\mathrm{OH})_3\) and \(\mathrm{H}_2 \mathrm{SO}_4\), and more \(\mathrm{H}_2 \mathrm{SO}_4\) is needed than is available (0.500 mol available vs 0.750 mol needed), \(\mathrm{H}_2 \mathrm{SO}_4\) is the limiting reactant.

Step 4: Determine Moles of Product Formed

Based on the limiting reactant (\(0.500 \text{ mol} \) \(\mathrm{H}_2 \mathrm{SO}_4\)), we use the reaction stoichiometry. From the balanced equation, 3 moles of \(\mathrm{H}_2 \mathrm{SO}_4\) produce 1 mole of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\). Thus, \(\frac{0.500}{3} = 0.167 \text{ mol} \) of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) can be formed.

Step 5: Calculate Excess Reactant Remaining

Use up 0.333 moles of \(\mathrm{Al} (\mathrm{OH})_3\) (needed to react with 0.500 moles \(\mathrm{H}_2 \mathrm{SO}_4\)), leaving \(0.500 - 0.333 = 0.167 \text{ mol} \) of \(\mathrm{Al} (\mathrm{OH})_3\) unreacted.

Key Concepts

Mole Ratio

In a chemical reaction, the mole ratio signifies the proportion of moles between different reactants and products as determined by a balanced chemical equation. This is essential for converting between amounts of reactants and products. When analyzing the reaction between aluminum hydroxide and sulfuric acid, the equation tells us:

• 2 moles of \( \mathrm{Al} (\mathrm{OH})_3 \)
• react with 3 moles of \( \mathrm{H}_2 \mathrm{SO}_4 \)
• producing 1 mole of \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \)

This 2:3:1 mole ratio is a quantitative measurement that guides how much of each substance is needed or produced. It is essential to calculate the correct amounts to avoid having too little or too much of a reactant when performing reactions. This step is crucial before moving on to determine which reactant will limit product formation.

Stoichiometry

Stoichiometry involves using mole ratios from balanced chemical equations to determine relationships between reactants and products. It is the central framework for understanding and performing calculations in chemistry. In this problem, stoichiometry is used to calculate how many moles of each reactant are necessary for the reaction to occur without leaving any reactant in excess.
For example, if 0.500 moles of \( \mathrm{Al} (\mathrm{OH})_3 \) are present, stoichiometry tells us that 0.750 moles of \( \mathrm{H}_2 \mathrm{SO}_4 \) are needed to fully react with it. Conversely, utilizing 0.500 moles of \( \mathrm{H}_2 \mathrm{SO}_4 \) requires only 0.333 moles of \( \mathrm{Al} (\mathrm{OH})_3 \) to be used completely. This way of calculating helps determine which reactant is the limiting one and therefore limits the amount of product that can be formed.

Excess Reactant

In any chemical reaction, the reactant that remains after the complete consumption of the limiting reactant is termed the excess reactant. It is vital to know how much of this reactant remains because it affects the reaction's efficiency and cost.
Using our reaction, we find that \( \mathrm{H}_2 \mathrm{SO}_4 \) is the limiting reactant, leaving \( \mathrm{Al} (\mathrm{OH})_3 \) as the excess reactant. With 0.500 moles of \( \mathrm{Al} (\mathrm{OH})_3 \), only 0.333 moles are required when \( \mathrm{H}_2 \mathrm{SO}_4 \) runs out. Thus, 0.167 moles remain unreacted.

This calculation is necessary for understanding the actual consumption of materials in a reaction and optimizing for future reactions.

Chemical Reactions

A chemical reaction involves the transformation of reactants into products through the breaking and forming of bonds. It is a process fundamental to creating new materials and substances. The given reaction is balanced, meaning that the number of atoms for each element is conserved across the reactants and products. Aluminum hydroxide \( (\mathrm{Al} (\mathrm{OH})_3) \) reacts with sulfuric acid \( (\mathrm{H}_2 \mathrm{SO}_4) \), resulting in the formation of aluminum sulfate \( (\mathrm{Al}_2(\mathrm{SO}_4)_3) \) and water \( (\mathrm{H}_2 \mathrm{O}) \).
Balanced chemical equations are crucial for determining the proportions in which chemicals react and the products yield. Not only do they show what substances are consumed and formed, but they also provide the necessary groundwork for the stoichiometric and limiting reactant calculations discussed earlier.