Limiting Reactant Procedure (Answer part c)

In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over: 2A +B-+A2B But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately: 2.8 met x SS = 1.4 mol A,B 1 A: 3.2 metB x ag = 3.2 mol A,B Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of AyB can be formed from the given amounts. 0.815 mol Previous Answers v Correct Correct answer is shown. Your answer 0.816 mol was either rounded differently or used a different number of significant figures than required for this part. Part B If you had excess aluminum, how many moles of aluminum chloride could be produced from 27.0 g of chlorine gas, Cl,? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) | 0.254 mol (Gia) Praious anemone v Correct Correct answer is shown. Your answer 0.250 mol was either rounded differently or used a different number of significant figures than required for this part. By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces tt Part C Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) + 3Cl: (g)-22 AlCl (s) What is the maximum mass of aluminum chloride that can be formed when reacting 22.0 g of aluminum with 27.0 g of chlorine? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) FAM .- Value Units