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 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?

Short answer

The limiting reactant is H\(_{2}\)SO\(_{4}\). Under these conditions, 0.167 moles of Al\(_{2}\)(SO\(_{4}\))\(_{3}\) can be formed, and 0.167 moles of Al(OH)\(_{3}\) remain after the completion of the reaction.

Step-by-step solution

Write the balanced equation

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

Determine the mole ratios

Here, we can see that the ratio of moles of Al(OH)\(_{3}\) to moles of H\(_{2}\)SO\(_{4}\) is 2:3. This means that for every 2 moles of Al(OH)\(_{3}\), 3 moles of H\(_{2}\)SO\(_{4}\) are needed for the reaction to proceed to completion.

Find the limiting reactant

Compare the given amounts of the reactants to the stoichiometric ratio: $$ \frac{\text{moles of Al(OH)}_3}{\text{moles of H}_2\text{SO}_4} = \frac{0.500}{0.500} = 1 $$ Now compare this value to the stoichiometric ratio: $$ \frac{2}{3} \approx 0.67 $$ Since 1 > 0.67, we have more Al(OH)\(_{3}\) than required for the given amount of H\(_{2}\)SO\(_{4}\). Therefore, H\(_{2}\)SO\(_{4}\) is the limiting reactant, and Al(OH)\(_{3}\) is in excess.

Calculate the amount of Al\(_{2}\)(SO\(_{4}\))\(_{3}\) formed

We can now use the limiting reactant to determine the amount of product formed. According to the balanced equation, 3 moles of H\(_{2}\)SO\(_{4}\) produce 1 mole of Al\(_{2}\)(SO\(_{4}\))\(_{3}\). So for 0.500 moles of H\(_{2}\)SO\(_{4}\), we get: $$ \text{moles of Al}_{2}\text{(SO}_{4}\text{)}_{3} = \frac{1}{3} \times 0.500 = 0.167 \, \text{moles} $$

Determine the amount of excess reactant remaining

To find the amount of Al(OH)\(_{3}\) remaining after the reaction, we first need to find how much Al(OH)\(_{3}\) was consumed by the limiting reactant. According to the balanced equation, 2 moles of Al(OH)\(_{3}\) react with 3 moles of H\(_{2}\)SO\(_{4}\). So for 0.500 moles of H\(_{2}\)SO\(_{4}\), we have: $$ \text{moles of Al(OH)}_3\text{ consumed} = \frac{2}{3} \times 0.500 = 0.333\, \text{moles} $$ Now we can subtract the amount of Al(OH)\(_{3}\) consumed from the initial amount to find how much is remaining: $$ \text{moles of Al(OH)}_3\text{ remaining} = 0.500 - 0.333 = 0.167\, \text{moles} $$ So, the limiting reactant is H\(_{2}\)SO\(_{4}\), 0.167 moles of Al\(_{2}\)(SO\(_{4}\))\(_{3}\) can form under these conditions, and 0.167 moles of Al(OH)\(_{3}\) remain after the completion of the reaction.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction. Crucial to understanding stoichiometry is the concept of the mole, which is a unit for counting particles of matter, typically atoms, molecules, or ions.

When performing stoichiometric calculations, a balanced chemical equation is vital because it shows the proportional relationship between reactants and products, based on the law of conservation of mass. This fundamental principle states that in a chemical reaction, mass is neither created nor destroyed, so the mass of the reactants must equal the mass of the products. In practice, this means that if we know the quantity of one substance in a reaction, we can calculate the quantities of all the other substances.

• Identify the 'mole ratio' from the balanced equation.
• Use the mole ratio to convert between moles of reactant and moles of product.
• Apply the 'limiting reactant' concept to determine which reactant will be completely used up first, thus stopping the reaction and determining the maximum amount of product that can be formed.

Chemical Reaction

A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Reactants are substances initially present in a chemical reaction that are consumed during the reaction to make products. When writing a chemical equation, reactants are listed on the left-hand side, while products are listed on the right-hand side.

• A chemical reaction can involve elements, compounds, or both.
• Chemical equations must be balanced, meaning the number of atoms for each element in the reactants must equal the number of atoms in the products.

The reaction between aluminum hydroxide and sulfuric acid is an example of a double displacement reaction. This involves two compounds swapping parts to form two new compounds. Understanding the type of reaction helps predict the products and write a balanced chemical equation, which is the starting point for any stoichiometric calculation.

Mole Ratio

The mole ratio, derived from the coefficients of a balanced chemical equation, is a critical aspect of stoichiometry. It indicates the proportions of reactants and products that interact in a reaction. For example, in our chemical equation, the mole ratio between \(\mathrm{Al(OH)}_{3}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 2:3, which means two moles of aluminum hydroxide react with three moles of sulfuric acid to produce products.

When solving stoichiometry problems, you use these ratios to convert between moles of one substance to moles of another. The mole ratio is essentially a conversion factor that bridges the gap between reactants and products. This provides a clear pathway to answer questions about the quantities of substances before and after a chemical reaction.