91Ó°ÊÓ

Stoichiometry is a fundamental concept in chemistry that allows us to calculate the quantities of reactants and products involved in chemical reactions. It relies on the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. This means that the mass of the reactants must equal the mass of the products. In our exercise, stoichiometry helps us determine how much sulfuric acid (\(\text{H}_2\text{SO}_4\)) is needed to fully react with a given mass of lanthanum hydroxide (\(\text{La(OH)}_3\)) in the neutralization reaction.When solving for stoichiometry, understanding the balanced chemical equation is crucial. The balanced equation tells us the mole ratio of reactants to products. In this case, the balanced equation is \[2 \text{H}_2\text{SO}_4 + 2 \text{La(OH)}_3 \rightarrow \text{La}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O}\]. From this, we see that 2 moles of \(\text{H}_2\text{SO}_4\) react with 2 moles of \(\text{La(OH)}_3\). Hence, for every mole of \(\text{La(OH)}_3\), we need an equal amount of sulfuric acid.Using stoichiometry, a student can calculate how many millimoles of \(\text{H}_2\text{SO}_4\) are required if given the mass of \(\text{La(OH)}_3\). This involves calculating how many moles of \(\text{La(OH)}_3\) are present and then using the mole ratio from the balanced equation to find the corresponding amount of \(\text{H}_2\text{SO}_4\) needed.

Chemical equations are symbolic expressions that represent chemical reactions in terms of chemical symbols and formulas. A chemical equation consists of reactants on the left side and products on the right side, with an arrow pointing from reactants to products, indicating the direction of the reaction.In our exercise, we wrote the chemical equation for the neutralization of \(\text{H}_2\text{SO}_4\) with \(\text{La(OH)}_3\). Neutralization reactions typically involve an acid and a base, forming water and a salt as products. Here, the acid is sulfuric acid (\(\text{H}_2\text{SO}_4\)) and the base is lanthanum hydroxide (\(\text{La(OH)}_3\)), resulting in lanthanum sulfate and water.A balanced chemical equation, like the one provided \[2 \text{H}_2\text{SO}_4 + 2 \text{La(OH)}_3 \rightarrow \text{La}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O}\], shows that the number of atoms for each element is the same on both sides. Balancing equations is critical in stoichiometry because it ensures that the law of conservation of mass is maintained. Each coefficient in the equation represents the number of moles of a substance involved in the reaction, providing a direct relationship between reactants and products.

Molar mass calculations are essential for converting between the mass of a substance and the amount in moles. The molar mass of a compound is the sum of the molar masses of its constituent elements, which can be found on the periodic table.For the compound \(\text{La(OH)}_3\), we calculated its molar mass as follows:

• Lanthanum (La): 138.91 g/mol
• Oxygen (O): 16.00 g/mol
• Hydrogen (H): 1.01 g/mol

Using these values, the molar mass of \(\text{La(OH)}_3\) is computed by summing the mass of lanthanum with three times the sum of the masses of oxygen and hydrogen:\[138.91 + 3 \times (16.00 + 1.01) = 189.94\,\text{g/mol}\]Knowing the molar mass is crucial for calculating moles from a given mass. In the exercise, given 0.8457 grams of \(\text{La(OH)}_3\), the moles are determined by dividing this mass by the molar mass:\[\frac{0.8457\,\text{g}}{189.94\,\text{g/mol}} = 0.00445\,\text{mol}\]Molar mass calculations thus bridge the gap between the tangible mass measurements we can make in a lab and the abstract concept of mixing chemical formulas in accurate ratios for desired reactions.