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?

Short answer

The limiting reactant is H2SO4. Under these conditions, 0.167 moles of Al2(SO4)3 can form, and 0.167 moles of excess reactant, Al(OH)3, remain after the reaction is complete.

Step-by-step solution

Determine the mole ratio of the reactants in the balanced equation

From the balanced chemical equation, we can see the mole ratio between the reactants Al(OH)3 and H2SO4 is 2:3. This means that for this reaction: 2 moles of Al(OH)3 react with 3 moles of H2SO4.

Determine the number of moles of each reactant given

We are given the number of moles for each reactant: 0.500 moles of Al(OH)3 and 0.500 moles of H2SO4.

Determine the limiting reactant

To determine the limiting reactant, we can calculate the relative amount of each reactant by dividing the number of moles of the reactant by its coefficient in the balanced equation: For Al(OH)3: \(0.500 \div 2 = 0.250\) For H2SO4: \(0.500 \div 3 = 0.167\) Since the relative amount of H2SO4 (0.167) is less than the relative amount of Al(OH)3 (0.250), H2SO4 is the limiting reactant.

Calculate the amount of product formed

Using the limiting reactant (H2SO4) and the mole ratio from the balanced equation (2:3 for Al(OH)3 and H2SO4), we can determine the moles of product (Al2(SO4)3) formed: For every 3 moles of H2SO4, 1 mole of Al2(SO4)3 is formed. So, \(0.500 \,\textit{moles}\,\textit{H2SO4} \times \frac{1 \, \textit{mole}\,\textit{Al}_2(\textit{SO}_4 )_3}{3 \, \textit{moles}\,\textit{H2SO4}} = 0.167 \, \textit{moles}\,\textit{Al}_2(\textit{SO}_4 )_3 \) Hence, 0.167 moles of Al2(SO4)3 can form under these conditions.

Determine the moles of excess reactant remaining

To find the remaining moles of the excess reactant (Al(OH)3), we will first determine the moles of Al(OH)3 that reacted with the limiting reactant (H2SO4): \(0.167 \, \textit{moles}\,\textit{Al}_2(\textit{SO}_4 )_3 \times \frac{2 \, \textit{moles}\,\textit{Al}(\textit{OH})3}{1 \, \textit{mole}\,\textit{Al}_2(\textit{SO}_4 )_3} = 0.333 \, \textit{moles}\,\textit{Al}(\textit{OH})_3 \textit{reacted}\) Now, subtract the moles of reacted Al(OH)3 from the initial moles of Al(OH)3 to find the remaining moles: \(0.500 \,\textit{moles}\,\textit{Al}(\textit{OH})_3 - 0.333 \,\textit{moles}\,\textit{Al}(\textit{OH})_3 = 0.167 \,\textit{moles}\,\textit{Al}(\textit{OH})_3 \) Hence, 0.167 moles of Al(OH)3 remain after the reaction is complete.

Key concepts

Stoichiometry

Understanding stoichiometry is crucial for determining the amounts of substances involved in chemical reactions. This field of chemistry is like a recipe, providing the proportions of reactants needed to produce a specific amount of product.

Imagine you're baking cookies - each ingredient must be added in the right amount. In stoichiometry, we use the 'mole' as the measuring unit, similar to using a cup or tablespoon in baking. By knowing the balanced chemical equation, we can calculate the moles needed or produced of each substance. The equation given in the exercise demonstrates stoichiometry in action, where a specific ratio of reactants yields a set amount of product.

In the exercise, the balanced equation was used to find the limiting reactant and predict the amount of product formed, exemplifying stoichiometry's application. Key to effectively using stoichiometry is starting with a balanced chemical equation, which ensures the law of conservation of mass is followed, meaning all atoms in the reactants will be accounted for in the products.

Chemical Reactions

Chemical reactions are processes where reactants transform into products. Like a dance, atoms in reactants break apart and rearrange to form new substances. It is essential to understand the types of chemical reactions, such as combination, decomposition, single-replacement, double-replacement, and combustion, to predict the products of a reaction.

The exercise showcases a double-replacement reaction where the anions and cations of two compounds switch places. This type of reaction often occurs in aqueous solutions, like the sulfuric acid and aluminum hydroxide reaction provided. Recognizing reaction types helps chemists control and predict the outcomes, ensuring the correct products are formed in the desired quantities. Furthermore, identifying the limiting reactant, the substance that runs out first and stops the reaction, is a crucial aspect of understanding complete chemical reactions.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we experience. One mole is defined as the amount of a substance that contains exactly 6.022 x 10²³ particles, which is Avogadro's number. This concept is akin to a dozen eggs - no matter the size of the eggs, a dozen is always twelve. Similarly, a mole of any substance will contain the same number of entities.

In the context of the exercise, understanding the mole concept allowed the calculation of product formed and excess reactant remaining after the chemical reaction. By using the mole ratios provided in the balanced chemical equation, students can convert between moles of different substances. This concept ensures accurate predictions of product amounts and helps in identifying the limiting reactant in a reaction.