91Ó°ÊÓ

First, we need to determine the limiting reactant by comparing the amount of each reactant to the stoichiometry given in the balanced chemical equation. In this case, for every 1 mole of A, we need 2 moles of B. Divide the number of moles of each reactant by their respective stoichiometric coefficients: \(A: \frac{16}{1} = 16\) \(B: \frac{20}{2} = 10\) The smaller number indicates the limiting reactant, which in this case is B.

Now that we have determined the limiting reactant, we can calculate the theoretical yield of product C based on the stoichiometry of the balanced chemical equation. From the balanced equation, we know that 1 mole of A reacts with 2 moles of B to form 1 mole of C. Since B is the limiting reactant, we can use the moles of B to determine the maximum moles of C that can be produced. Use the coefficients from the balanced equation to convert moles of B to moles of C: \(moles \ of \ C = (\frac{1 \ mol \ of \ C}{2 \ mol \ of \ B}) \times moles \ of \ B\) Plug in the number of moles we had at the beginning (20 moles of B): \(moles \ of \ C = (\frac{1}{2}) \times 20\) Calculate the moles of C: \(moles \ of \ C = 10\) So, the theoretical yield for product C in this reaction is 10 moles. This quantity depends on the limiting reactant, which in this case is B.