Notes Chapter 9 Limiting Reagent Sample Problems Page 1 Problem 1: Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas. Suppose that 6.70 Na reacts with 3.20 Cl 2. A. What is the limiting reagent? B. How many es of NaCl are produced? 1. Write the e amounts under their respective places. 6.7 3.2 from the Na to the Cl 2, or work from the Cl 2 to the Na. The choice is yours.) (I choose to work from Na to Cl 2 ) 6.7 3.2 6.7 Na 1 Cl 2 = 3.35 of Cl 2 needed for this reaction to take place. 2 Na 3. Now determine if the problem gave you enough Cl 2 for this reaction to take place. No they didn t. We need 3.35 es of Cl 2, and they only gave us 3.2 es of Cl 2. Since we don t have enough Cl 2, it must be our limiting reagent. 4. So, the limiting reagent for this reaction is the Cl 2. reagent over to what the problem wants to know (How many es of NaCl are 6.7 3.2? 3.2 Cl 2 2 NaCl = 6.40 NaCl are produced 1 Cl 2
Notes Chapter 9 Limiting Reagent Sample Problems (continued) Page 2 Problem 2: The equation for the complete combustion for ethane (C 2 H 4 ) is written below. If 2.70 C 2 H 4 is reacted with 6.30 O 2, A. Identify the limiting reagent. B. Calculate the es of water produced. 1. Write the e amounts under their respective places. 2.70 6.30 from the C 2 H 4 to the O 2, or work from the O 2 to the C 2 H 4. The choice is yours.) (I choose to work from C 2 H 4 to O 2 ) 2.70 6.30 2.70 C 2 H 4 3 O 2 = 8.1 of O 2 needed for this reaction to take place. 1 C 2 H 4 3. Now determine if the problem gave you enough O 2 for this reaction to take place. No they didn t. We need 8.1 es of O 2, and they only gave us 6.3 es of O 2. Since we don t have enough O 2, it must be our limiting reagent. 4. So, the limiting reagent for this reaction is the O 2. reagent over to what the problem wants to know (How many es of H 2 O are 2.70 6.30? 6.3 O 2 2 H 2 O = 4.2 H 2 O are produced 3 O 2
Notes Chapter 9 Limiting Reagent Sample Problems (continued) Page 3 Problem 3: The equation for the incomplete combustion for ethane (C 2 H 4 ) is written below. If 2.70 C 2 H 4 is reacted with 6.30 O 2, C. Identify the limiting reagent. D. Calculate the es of water produced. C 2 H 4 + 2 O 2 2 CO + 2 H 2 O (notice we have different products.) 1. Write the e amounts under their respective places. C 2 H 4 + 2 O 2 2 CO + 2 H 2 O 2.70 6.30 from the C 2 H 4 to the O 2, or work from the O 2 to the C 2 H 4. The choice is yours.) C 2 H 4 + 2 O 2 2 CO + 2 H 2 O (I choose to work from C 2 H 4 to O 2 ) 2.70 6.30 2.70 C 2 H 4 2 O 2 = 5.4 of O 2 needed for this reaction to take place. 1 C 2 H 4 3. Now determine if the problem gave you enough O 2 for this reaction to take place. Yes they did. We only need 5.4 es of O 2, and they gave us 6.3 es of O 2. Since the problem gave us more than enough O 2, the other guy (the C 2 H 4 ) must be our limiting reagent. 4. So, the limiting reagent for this reaction is the C 2 H 4. reagent over to what the problem wants to know (How many es of H 2 O are 2.70? 2.7 C 2 H 4 2 H 2 O = 5.4 H 2 O are produced 1 C 2 H 4
Notes Chapter 9 Limiting Reagent Sample Problems (continued) Page 4 Problem 4: A. What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? B. What is the maximum number of grams of Cu 2 S that can be formed? 2 Cu + S Cu 2 S 1. Write the e amounts under their respective places. 2 Cu + S Cu 2 S 80.0 g 25.0 g from the Cu to the S, or work from the S to the Cu. The choice is yours.) 2 Cu + S Cu 2 S (I choose to work from Cu to S) 80.0 g 25.0 g 80.0 g Cu 1 Cu 1 S 32 g S = 20.0 g S needed for this 64 g Cu 2 Cu 1 S reaction to take place. 3. Now determine if the problem gave you enough S for this reaction to take place. Yes they did. We only need 20.0 g of S, and they gave us 25.0 g of S. Since the problem gave us more than enough S, the other guy (the Cu) must be our limiting reagent. 4. So, the limiting reagent for this reaction is the Cu. reagent over to what the problem wants to know (How many grams of Cu 2 S are 2 Cu + S Cu 2 S (Work from the over to the Cu 2 S.) 80.0 g? g 80.0 g Cu 1 Cu 1 Cu 2 S 160 g Cu 2 S = 100 g of Cu 2 S produced 64 g Cu 2 Cu 1 Cu 2 S
Notes Chapter 9 Limiting Reagent Sample Problems (continued) Page 5 Problem 5: Hydrogen gas can be produced in the laboratory by the reaction of magnesium metal with hydrochloric acid. A. Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg. B. How many grams of hydrogen can be produced when 6.00 g HCl is added to 5.00 g Mg? Mg + 2 HCl MgCl 2 + H 2 1. Write the e amounts under their respective places. Mg + 2 HCl MgCl 2 + H 2 5.0 g 6.0 g from the Mg to the HCl, or work from the HCl to the Mg. The choice is yours.) Mg + 2 HCl MgCl 2 + H 2 ( I choose to work from Mg to HCl) 5.0 g 6.0 g 5.0 g Mg 1 Mg 2 HCl 36.5 g HCl = 15.2 g HCl needed for 24 g Mg 1 Mg 1 HCl this reaction to take place. 3. Now determine if the problem gave you enough HCl for this reaction to take place. No they didn t. We need 15.2 g of HCl, and they only gave us 6.0 g of HCl. Since we don t have enough HCl, it must be our limiting reagent. 4. So, the limiting reagent for this reaction is the HCl. reagent over to what the problem wants to know (How many grams of H 2 are Mg + 2 HCl MgCl 2 + H 2 ( I choose to work from Mg to HCl) 6.0 g? g 6.0 g HCl 1 HCl 1 H 2 2 g H 2 = 0.16 g H 2 are produced. 36.5 g HCl 2 HCl 1 HCl