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Balancing chemical equations is a critical first step in any chemical reaction analysis. In a balanced equation, the number of atoms of each element is the same on both sides of the equation, ensuring the law of conservation of mass is upheld. The given chemical reaction begins unbalanced: \( \mathrm{Al(OH)}_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + \mathrm{H}_{2} \mathrm{O} \). As a first step, balance the aluminum, hydroxide, hydrogen, sulfur, and oxygen atoms.

• Aluminum (\(\mathrm{Al}\)): Add a coefficient of 2 before \(\mathrm{Al(OH)}_{3}\) to balance the 2 \(\mathrm{Al}\) atoms on the right.
• Sulfate (\(\mathrm{SO}_{4}\)): Note that 2 \(\mathrm{Al}\) in \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) implies 3 sulfate groups; therefore, balance by placing a coefficient of 3 before \(\mathrm{H}_{2} \mathrm{SO}_{4}\).
• Water (\(\mathrm{H}_{2} \mathrm{O}\)): With hydrogen balanced across water molecules, ensure 6 \(\mathrm{H}_{2} \mathrm{O}\) generates 12 hydrogen atoms, matching those paired within \(\mathrm{H}_{2} \mathrm{SO}_{4}\).

The balanced equation is: \[ 2 \mathrm{Al(OH)}_{3} + 3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + 6 \mathrm{H}_{2} \mathrm{O} \] Balancing the chemical equation ensures all reactants and products are correctly proportioned, guiding the stoichiometry calculations that follow.

Moles form the bridge between the atomic world and the real-world quantities like mass and volume. For this exercise, determining the moles of reactants allows us to find the limiting reactant and calculate the products formed. Start with sulfuric acid where you convert the given volume into liters: 25.0 mL = 0.0250 L. Using molarity \( (M) \), apply the formula: \[ \text{moles of } \mathrm{H}_{2} \mathrm{SO}_{4} = \text{Molarity} \times \text{Volume in L} = 0.500 \times 0.0250 = 0.0125 \text{ moles}\] Next, calculate the moles of aluminum hydroxide \( (\mathrm{Al(OH)}_{3}) \) using its mass and molar mass (78.00 g/mol): \[ \text{moles of } \mathrm{Al(OH)}_{3} = \frac{1.00}{78.00} = 0.0128 \text{ moles}\] These calculations set the stage for determining which reactant is limiting and what quantities of products can be expected.

Stoichiometry involves using balanced chemical equations to predict the quantities of reactants used and products formed in a reaction. It's the quantitative backbone of chemistry. With a balanced equation and known moles, identify the limiting reactant. According to the equation: \[ 2 \mathrm{Al(OH)}_{3} + 3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + 6 \mathrm{H}_{2} \mathrm{O} \] From the stoichiometric coefficients, 2 moles of \(\mathrm{Al(OH)}_{3}\) reacts with 3 moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Compute the moles ratio:

• \(\mathrm{Al(OH)}_{3}\): \(\frac{0.0128}{2} = 0.0064\)
• \(\mathrm{H}_{2} \mathrm{SO}_{4}\): \(\frac{0.0125}{3} = 0.00417\)

The smaller ratio indicates \(\mathrm{H}_{2} \mathrm{SO}_{4}\) as the limiting reactant, controlling the reaction's yield. Using this, calculate the product's moles: \[ \text{From 3 moles of } \mathrm{H}_{2} \mathrm{SO}_{4}, \text{ produce } 6 \mathrm{H}_{2} \mathrm{O}\text{ thus } 0.0125 \times 2 = 0.0250 \text{ moles of } \mathrm{H}_{2} \mathrm{O}\] Finally, convert moles of water into mass using its molar mass (18 g/mol): \[ 0.0250 \times 18 = 0.450 \text{ g of } \mathrm{H}_{2} \mathrm{O} \] Stoichiometry helps you predict the maximum amount of product formed, given the limiting constraints in the reactants.