EXCESS QUANTITY & LIMITING QUANTITY

Transcription

1 EXCESS QUANTITY & LIMITING QUANTITY A balanced equation describes what should happen in a chemical reaction. However, the IDEAL conditions necessary for the reaction to take place may not be present. (ie. pressure, temperature, concentration, etc.) Reactions are frequently carried out where a reactant is present in excess. We may do this because: i) Deliberately adding an excess of one reactant makes sure the second one completely reacts. ii) Unavoidably having a reactant in excess because a limited amount of the other one is present. Ie. One reactant is the EXCESS QUANTITY and some of it will be left over after the reaction has come to completion. The second reactant is entirely used up and is the LIMITING REACTANT. Example: The reaction equation for making bicycles might be written: 1 seat + 2 wheels bicycle If a factory has (reacts) 4 wheels and 3 seats with sufficient other parts, (a) which reactant will be the limiting quantity and (b) how much of which reactant will be in excess (left unreacted)? Solution: (a) 3 seats will make 3 bicycles; but 4 wheels will only make 2 bicycles. Therefore, only 2 bicycles can be made. Since the number of wheels LIMITS the amount of product that can be made, the wheels will be the quantity. (b) Since 4 wheels will react with only 2 seats, the will be the excess quantity, and seat will be left over. Example 1: When 4.50 g of H 2 is reacted with 140 g Cl 2, how many grams of which reactant will be left over (IN EXCESS)? Step 1: Get a balanced equation for the reaction. H 2(g) + Cl 2(g) 2 HCl (g) Step 2: To find which reactant is in excess, calculate how many grams of chlorine would be required to react with 4.50 g of hydrogen gas g H 2 x 1 mole H 2 x 1 mol Cl 2 x 70.9 g of Cl 2 = 160 g Cl g H 2 1 mole H 2 1 mol Cl grams of Cl 2 gas would be required to react with 4.50 g of H 2 gas. Since there are only 140 g of Cl 2 gas, not all 4.50 g of H 2 gas can react. Therefore, is the limiting quantity and is the excess quantity.

2 However, to calculate how much of the excess quantity is left over, we must find out how much of the second reactant (H 2 gas) would react with the limiting reactant (140 g Cl 2 ). 140 g Cl 2 x 1 mole Cl 2 x 1 mole H 2 x 2.0 g H 2 = 3.95 g H g Cl 2 1 mole Cl 2 1 mole H 2 Only 3.95 g of H 2 would be needed to react with 140 g of Cl g H 2 (starting amount) g H 2 (consumed in reaction) 0.55 g H 2 (in excess, left over) Solution: 0.55 g H 2 gas would be in excess when 140 g Cl 2 gas reacts with 4.50 g H 2 gas. Example 2: The equation below tells us that 2 MOLES of Potassium permanganate reacts with 5 MOLES of Hydrosulfuric acid and 3 MOLES of Sulfuric acid to produce 1 MOLE of Potassium sulfate, 2 MOLES of Manganese sulfate, 8 MOLES of water and 5 MOLES of Sulphur. 2 KMnO H 2 S + 3 H 2 SO 4 K 2 SO MnSO H 2 O + 5 S If 48.0 g of KMnO 4 are reacted with 24.6 g of H 2 S and sufficient sulphuric acid, how much of which reactant will be left over? Step 1: Calculate the number of grams of KMnO 4 that would react with 24.6 g of H 2 S first g H 2 S x 1 mole H 2 S x 2 moles KMnO 4 x g KMnO 4 = 45.6 g KMnO g H 2 S 5 moles H 2 S 1 mole KMnO 4 That is, 45.6 g of KMnO 4 will react with 24.6 g H 2 S. (a) Will there be enough KMnO 4 to react with all the H 2 S? (b) Do we have to calculate how much H 2 S reacts with 48.0 g of KMnO 4 as well? Why or why not? (c) What is the limiting quantity? Explain why.

3 (d) What reactant is in excess and by how much? Show your work. EXERCISE A For the following chemical reactions, determine which reactant is in EXCESS and which reactant is LIMITING. Determine how much of the EXCESS reactant will be left over at the end of the reaction. Use the following balanced equation for questions #1 and #2. C (g) + O 2(g) CO 2(g) 1. A reaction of 2.19 moles of solid carbon with 3.48 moles of oxygen gas to form carbon dioxide. 2. A reaction of 2.00 g of solid Carbon with 5.00 g of Oxygen gas to form carbon dioxide. Use the following balanced chemical equation for questions #3 and #4. N 2(g) + 3 H 2(g) 2 NH 3(g) 3. A reaction of 1.4 moles of Nitrogen gas with 1.8 moles of Hydrogen gas to form Ammonia. 4. A reaction of 50.4 g of Nitrogen gas with 3.90 g of Hydrogen gas to form Ammonia.

4 CALCULATING AMOUNT OF PRODUCT FORMED WHEN ONE OF THE REACTANTS IS IN EXCESS Example 3: How many grams of HCl will be formed when 4.50 g of H 2 react with 140 g of Cl 2. H 2(g) + Cl 2(g) 2 HCl (g) Step 1: Calculate the grams of HCl formed by the given amount of each reactant. 140 g Cl 2 x 1 mole Cl 2 x 2 mole HCl x 36.5 g HCl = 144 g HCl 70.9 g Cl 2 1 mole Cl 2 1 mole HCl 144 g of HCl would be produced if 140 g of Cl 2 reacted with sufficient H g H 2 x 1 mole H 2 x 2 moles HCl x 36.5 g HCl = 164 g HCl 2.0 g H 2 1 mole H 2 1 mole HCl 164 g of HCl would be produced if 4.5 g of H 2 reacted with sufficient Cl 2. (a) What is the maximum amount of HCl that can be produced? (b) What is the limiting quantity? Explain why. (c) What is the excess quantity? Explain why. EXERCISE B 1. What mass of HCl (g) is produced when 9.50 g of H 2(g) and g of Cl 2(g) are reacted? H 2(g) + Cl 2(g) 2 HCl (g) 2. What mass of S is produced when 15.6 g of KMnO 4, 7.95 g of H 2 S and 15.3 g of H 2 SO 4 are reacted? 2 KMnO H 2 S + 3 H 2 SO 4 K 2 SO MnSO H 2 O + 5 S

5 3. If g of Aluminum are reacted with g of Copper(II)sulfate, CuSO 4, then Aluminum sulfate, Al 2 (SO 4 ) 3, and Copper are formed according to the following balanced chemical equation: 2 Al (s) + 3 CuSO 4(aq) Al 2 (SO 4 ) 3(aq) + 3 Cu (s) (a) Which reactant will be in excess? (b) Calculate the mass of the excess reactant which will remain after the reaction is completed. (c) Calculate the mass of each product formed in the reaction, using the LIMITING reactant.

6 PERCENT YIELD Sometimes 100% of the expected amount of products cannot be achieved in a reaction because - the reactants don t all react or - some product is lost during procedures (filtration, transfer, crystallization, etc.) The term Percent Yield is used to express how much (in percent) is actually obtained. % YIELD = AMOUNT OF PRODUCT RECOVERED X 100% AMOUNT OF PRODUCT EXPECTED % YIELD = ACTUAL YIELD X 100% THEORETICAL YIELD Example 1: In the reaction: HBrO HBr 3 H 2 O + 3 Br grams of HBrO 3 was reacted with excess HBr and 26.5 grams of Br 2 was produced. What was the % yield of this reaction. (26.5 g of Br 2 is the ACTUAL YIELD) Step 1: Find the amount of Br 2 expected to be produced (THEORETICAL YIELD) when 10.0 grams of HBrO 3 reacts. Theoretical Yield: 10.0 g HBrO 3 x 1 mole HBrO 3 x 3 moles Br 2 x g Br 2 = 37.2 g Br g HBrO 3 1 mole HBrO 3 1 mole Br 2 Step 2: % Yield = Actual Yield x 100% = 26.3 g Br 2 x 100% = 70.7 % yield Theoretical Yield 37.2 g Br 2 For every 100 grams of product predicted, only 70.7 grams of product is actually formed. Example 2: In the reaction: C 3 H 6 Cl KSCN C 3 H 6 (SCN) KCl The percent yield is 80%. How many grams of C 3 H 6 (SCN) 2 will be produced if 3.00 grams of C 3 H 6 Cl 2 is reacted with excess KSCN? Step 1: Calculate the theoretical yield for C 3 H 6 (SCN) 2 expected to be produced when 3.00 grams of C 3 H 6 Cl 2 reacts g C 3 H 6 Cl 2 x 1 mole C 3 H 6 Cl 2 x 1 mole C 3 H 6 (SCN) 2 x g C 3 H 6 (SCN) 2 = 4.19 g g C 3 H 6 Cl 2 1 mole C 3 H 6 Cl 2 1 mole C 3 H 6 (SCN) 2 Step 2: % Yield = Actual Yield of C 3 H 6 (SCN) 2 x 100% = 80% yield 4.19 g C 3 H 6 (SCN) 2

7 Rearrange formula and solve for the Actual Yield of C 3 H 6 (SCN) 2 : Actual Yield: 80% x 4.19 g C 3 H 6 (SCN) 2 = 3.35 g C 3 H 6 (SCN) 2 100% EXERCISE C 1. Pentane, C 5 H 12 burns according to the reaction: C 5 H 12 (l) + 8 O 2(g) 5 CO 2(g) + 6 H 2 O (l) (a) If 150 g of pentane was reacted with excess oxygen and 392 g of carbon dioxide was produced, what was the % yield of the reaction? (b) If % yield is 60.2%, how many grams of CO 2 will be produced when 16.5 g of pentane is reacted? 2. If 12.5 g of Copper are reacted with an excess of Chlorine, then 25.4 g of Copper (II) chloride, CuCl 2, are obtained. Calculate the theoretical yield and percent yield. Cu (s) + Cl 2(g) CuCl 2(s) 3. If 6.57 g of Iron are reacted with an excess of Hydrochloric acid, HCl, then Hydrogen gas and g of Iron (III) chloride are obtained. Calculate the theoretical yield and percent yield. 2 Fe (s) + 6 HCl (aq) 3 H 2(g) + 2 FeCl 3(aq)