1. Limiting Reagent Problem

2. Calculate the mass of aluminum chloride that can be produced from 20.0 g of aluminum and 30.0 g of chlorine gas. Step 1: Write the balanced equation for the reaction. 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s)

3. Step 2: Calculate the number of moles of each reactant. (Using n=m/M) 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s) m 20.0 g 30.0g M 27.0g/mol 2(35.5) =71.0g/mol n = 0.741 mol = 0.423 mol Cl 2 20.0g 27.0g/mol 30.0g 71.0 g/mol

4. Step 3: Use the balanced equation to determine the limiting reagent. 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s) 2 Al 2 AlCl 3 = 0.742 mol x x= 0.741 mol AlCl 3 3 Cl 2 AlCl 3 = 0.423 mol x x= 0.282 mol AlCl 3 Cl 2 produces the smallest amount of AlCl 3. It will run out first, so it is the Limiting reagent. We use Cl 2 for the rest of the problem. Cl 2 will run out when it has produced n= 0.282 mol of AlCl 3

5. Step 4. Calculate the number of moles using the limited reagent. Moles of AlCl 3 = nM = 0.282mol(133.33g/mol) = 37.6 g of AlCl 3

6. Try these: p. 254 – 258 # 24, 25, 28ab, 30a p. 259 #3