# Please TURN IN C.7/C.10 worksheet Objectives: Determine the limiting reagent in a reaction. Calculate the percent yield of a product in a reaction Catalyst:

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Please TURN IN C.7/C.10 worksheet Objectives: Determine the limiting reagent in a reaction. Calculate the percent yield of a product in a reaction Catalyst: 1 Double Cheeseburger needs 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon For 5 Double Cheeseburgers how many units of each ingredient do I need? Fill in the final column below with how many complete burgers I can make: BunsPattiesCheeseBacon Burgers? 2448 44816 303248

A. The Concept of Limiting Reactants Stoichiometric mixture – N 2 (g) + 3H 2 (g)  2NH 3 (g)

Limiting reactant mixture A. The Concept of Limiting Reactants –N 2 (g) + 3H 2 (g)  2NH 3 (g)

For a Limiting reactant mixture the number of moles are not balanced to match the reaction equationN 2 (g) + 3H 2 (g)  2NH 3 (g) A. The Concept of Limiting Reactants –Limiting reactant is the reactant that runs out first –When the limiting reactant is exhausted, then the reaction stops

LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4 Na (s) + O 2(g)  2 Na 2 O (s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 6.74 g of Na 2 O 5.00g Na ( 1 mole Na ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 6.74 g of Na 2 O 23 g Na 4 mole Na 1 mol Na 2 O 19.38 g of Na 2 O 5.00g O 2 ( 1 mole O 2 ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 19.38 g of Na 2 O 32 g O 2 1 mole O 2 1 mol Na 2 O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4 Na (s) + O 2(g)  2 Na 2 O (s) The amount of O 2 used to make 6.74 g of Na 2 O is calculated by: 1.74 g of O 2 was used 5.00g Na ( 1 mole Na ) ( 1 mole O 2 )( 32 g O 2 ) = 1.74 g of O 2 was used 23 g Na 4 mole Na 1 mol O 2 The amount of oxygen (O 2 ) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

PRACTICE PROBLEMS 2 C 2 H 2 + 5 O 2  4 CO 2 + 2 H 2 O 1. How many moles of carbon dioxide could be produced from 220.0 g of C 2 H 2 and 545.0 g of O 2 ? 2.How many grams of CO 2 can be produced by the reaction of 35.5 grams of C 2 H 2 and 45.9 grams of O 2 ? 50.5 g 13.63 mol

C. Percent Yield Theoretical Yield – The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. The actual yield (amount produced) of a reaction is usually less than the maximum expected (theoretical yield). Percent Yield – The actual amount of a given product as the percentage of the theoretical yield.

Calcium carbonate is synthesized by heating,as shown in the following equation: CaO + CO 2  CaCO 3 What is the theoretical yield of CaCO 3 if 24.8 g of CaO is heated with 43.0 g of CO 2 ?What is the theoretical yield of CaCO 3 if 24.8 g of CaO is heated with 43.0 g of CO 2 ? What is the percent yield if 33.1 g of CaCO 3 is produced?What is the percent yield if 33.1 g of CaCO 3 is produced? Determine which reactant is the limiting and then decide what the theoretical yield is.

24.8 g CaO 1molCaO 56g CaO 1mol CO 2 1mol CaO 44 g CO 2 1molCO 2 = 19.5gCO 2 24.8gCaO  molCaO  mol CO 2  gCO 2 24.8 g CaO 1mol CaO 56g CaO 1molCaCO 3 1mol CaO 100g CaCO 3 1molCaCO 3 = 44.3 g CaCO 3 24.8gCaO  molCaO  mol CaCO 3  gCaCO 3 LR

CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO 3. We only produced 33.1 g. CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO 3. We only produced 33.1 g. Our percent yield is: Our percent yield is: Percent yield= 33.1 g CaCO 3 44.3 g CaCO 3 _____________ x 100 Percent yield = 74.7%

Percent Yield Questions 1. The electrolysis of water forms H 2 and O 2. 2H 2 O  2H 2 + O 2 What is the % yield of O 2 if 12.3 g of O 2 is produced from the decomposition of 14.0 g H 2 O? 2. Iron pyrites (FeS 2 ) reacts with oxygen according to the following equation: 4FeS 2 + 11O 2  2Fe 2 O 3 + 8SO 2 If 300 g of iron pyrites is burned in 200 g of O 2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?

1.The electrolysis of water forms H 2 & O 2. 2H 2 O  2H 2 + O 2 Give the percent yield of O 2 if 12.3 g O 2 is produced from the decomp. of 14 g H 2 O? Actual yield is given: 12.3 g O 2 Next, calculate theoretical yield 1 mol O 2 2 mol H 2 O x # g O 2 = 14.0 g H 2 O 12.43 g= 32 g O 2 1 mol O 2 x 1 mol H 2 O 18.02 g H 2 O x Finally, calculate % yield 12.3 g O 2 12.43 g O 2 = % yield = x 100%98.9%= actual theoretical x 100%

2. 4FeS 2 + 11O 2  2Fe 2 O 3 + 8SO 2 If 300 g of FeS 2 is burned in 200 g of O 2, 143 g Fe 2 O 3 results. % yield Fe 2 O 3 ? First, determine limiting reagent 2 mol Fe 2 O 3 11 mol O 2 x 200 g O 2 181.48 g Fe 2 O 3 = 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol O 2 32 g O 2 x 2 mol Fe 2 O 3 4 mol FeS 2 x # g Fe 2 O 3 = 300 g FeS 2 199.7 g Fe 2 O 3 = 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol FeS 2 119.97 g FeS 2 x 143 g Fe 2 O 3 181.48 g Fe 2 O 3 = % yield = x 100%78.8%= actual theoretical x 100%

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