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Theoretical yield is the maximum amount of a product that can be produced in a chemical reaction, calculated based on the stoichiometry of the reaction and the amount of reactants present. It assumes that the reaction goes to completion and that there are no side reactions or losses.

Theoretical yield depends on the limiting reactant because the limiting reactant determines how much product can be formed in a reaction. The limiting reactant is the reactant that gets completely consumed first and thus limits the amount of product that can be formed. In a balanced chemical reaction, the stoichiometric coefficients of the reactants and products determine the ratios in which the reactants react and the products are formed. When we have different amounts of reactants, the reactant present in the smallest amount based on its stoichiometric coefficient will be the limiting reactant. This will determine the maximum amount of product that can be formed in the reaction, i.e., the theoretical yield.

To calculate the theoretical yield, follow these steps: 1. Write a balanced equation for the reaction. 2. Determine the amount of each reactant in moles. 3. Calculate the "mole ratio" of reactants by dividing the number of moles of each reactant by their respective stoichiometric coefficients. The reactant with the lowest mole ratio is the limiting reactant. 4. Calculate the theoretical yield by multiplying the number of moles of the limiting reactant, the mole-to-mole ratio between the product and the limiting reactant, and the molar mass of the desired product. Here's an example to illustrate the process: Consider the reaction: \(2H_2 + O_2 \rightarrow 2H_2O\) Suppose we have 8 moles of \(H_2\) and 5 moles of \(O_2\). Step 1: The balanced equation is already provided. Step 2: Moles of \(H_2 = 8\); moles of \(O_2 = 5\). Step 3: Calculate the mole ratio: Mole ratio of \(H_2 = \frac{8}{2} = 4\); mole ratio of \(O_2 = \frac{5}{1} = 5\). In this case, the limiting reactant is \(H_2\) since it has a lower mole ratio. Step 4: Calculate the theoretical yield: The stoichiometric ratio between \(H_2O\) and \(H_2\) is 1:1, so the theoretical yield of \(H_2O\) is equal to the number of moles of the limiting reactant, \(H_2\). The theoretical yield in moles of \(H_2O = 8\). Convert this to mass using the molar mass of \(H_2O = 18.02 g/mol\): Theoretical yield of \(H_2O = 8 \times 18.02 = 144.16g\).