In this example, we will solve a limiting reagent problem using a limiting reagent table. The problem reads: "Given the reaction: 65 grams of N2H4 gas and 87.4 grams of N2O4 gas react to completion. Calculate the mass of NH3 gas that can be produced, the mass that remains of the excess reactant, and the percent yield if 64.5 grams of N2 gas is produced when the experiment is performed." We know this is a limiting reagent problem because we are given the amounts of two different reactants in the reaction. For a limiting reagent problem, we will calculate: 1. Starting moles (the number of reactions that can be performed with each of our reactants) 2. Change in moles 3. Moles of each at the end To calculate our starting moles, we can use dimensional analysis. We can convert 65 grams of N2H4 into moles of N2H4, which gives us 2.03 moles of N2H4. We can also convert our 87.4 grams of N2O4 into moles of N2O4 using its molar mass, which gives us 0.950 moles of N2O4. Next, we want to calculate how many reactions we can perform with each of these moles of reactants. We can also use dimensional analysis for this step. We have 2.03 moles of N2H4, and each reaction requires 2 moles of N2H4. So we can perform 1.01 reactions with N2H4. For N2O4, we do a similar calculation, converting moles of N2O4 into reactions using the coefficient in the balanced chemical equation. For every one reaction, we need one mole of N2O4, so we can perform 0.950 reactions. The reactant that can perform the fewest reactions is our limiting reagent. In this problem, N2O4 is our limiting reagent because it can produce the smallest number of reactions. It will run out first and thus limit how much product we can produce. To calculate our change in moles, we have to use the number of reactions that our limiting reagent can provide. So, to calculate our change in moles for N2H4, we can perform 0.950 reactions, and in each reaction, we lose 2 moles of N2H4. Therefore, we will lose 1.90 moles of N2H4. For N2O4 with the 0.950 reactions, we will use up 0.950 moles of N2O4. And for the product with 0.950 reactions, each reaction produces three moles of N2. So, we will produce 2.85 moles of N2. We have a positive change for the product. Since our problem doesn't ask us about H2O gas, we don't need to calculate the change in moles. To calculate our ending moles, we take our starting moles plus our change in moles. For N2H4, we start with 2.03 moles, lose 1.9 moles, and have 0.13 moles remaining. For N2O4, we started with 0.950 moles and lost 0.950 moles, so we have zero moles remaining of our limiting reagent. And for N2, we started with zero moles and produced 2.85 moles. So we have 2.85 moles of N2 when our reaction goes to completion. Now we can calculate the mass of N2 gas produced. We can convert our 2.85 moles of N2 gas into grams using its molar mass, and we'll get 79.8 grams of N2. This answers Part A. To answer Part B, we can use the ending moles of N2H4 gas. Since N2H4 gas is the excess reactant, we can convert the 0.13 moles into grams using its molar mass, and we'll get 4.20 grams of N2H4 remaining. This is the answer to Part B. To answer Part C, we're given an experimental value for how much product is produced: 64.5 grams of N2 gas. We've already calculated how much N2 gas should be able to be produced based on stoichiometry. The amount we were given in the problem that's actually produced when the experiment is performed is our actual yield. The amount we calculate using stoichiometry is our theoretical yield. So, to calculate percent yield, we take the actually produced 64.5 grams of nitrogen gas and divide it by the theoretically calculated 79.8 grams of nitrogen gas (which is what we calculated for Part A), and then multiply by 100 percent. We'll get 80.8 percent. This is the answer to Part C.