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Limiting Reagent and Percent Yield. Limiting Agent The limiting reagent limits or determines the amount of product that can be formed in a reaction. The.

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Presentation on theme: "Limiting Reagent and Percent Yield. Limiting Agent The limiting reagent limits or determines the amount of product that can be formed in a reaction. The."— Presentation transcript:

1 Limiting Reagent and Percent Yield

2 Limiting Agent The limiting reagent limits or determines the amount of product that can be formed in a reaction. The reaction occurs only until the limiting reagent is used up.

3 Limiting Agent The reactant that is not completely used up in a reaction is called the excess reagent. The amount of product formed in a reaction can be determined from the given amount of limiting reagent.

4 Determining the Limiting Reagent in a Reaction What is the limiting reagent when 80g Cu reacts with 25g S? – 2Cu(s) + S(s)  Cu 2 S(s) Step 1: convert to moles – 80g Cu x (1 mol Cu/63.5g Cu) = 1.26 mol Cu – 25g S x (1 mol S/32.1g S) = mol S

5 Determining the Limiting Reagent in a Reaction Step 2: Look at the mole ratio from the balanced equation: – There is 2 moles of Cu for 1 mole of S. Step 3: Find the amount of moles in the equation. – 1 mol of S =.779 mol S – 2 moles of Cu = 1.26 mol Cu x (1 mol S/2 mol Cu) = S Since the amount of sulfur needed is moles and the given amount is moles, this indicates that sulfur is in excess. Thus copper is the limiting reagent.

6 Determining the Limiting Reagent in a Reaction - Practice 23. If 2.70 mol C 2 H 4 is reacted with 6.30 mol O 2, identify the limiting reagent.

7 Determining the Limiting Reagent in a Reaction - Practice 24. Identify the limiting reagent when 6.00g HCl reacts with 5.00g Mg. – Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g)

8 Using a Limiting Reagent to Find the Quantity of a Product What is the maximum number of grams of Cu 2 S that can be formed from: – 2Cu(s) + S(s)  Cu 2 S(s) Step 1: Use moles of limiting reagent to calculate moles of what you are looking for. – 1.26 mol Cu (from previous problem) – 1.26 mol Cu x (1 mol Cu 2 S/2 mol Cu) = 0.63 mol Cu 2 S

9 Using a Limiting Reagent to Find the Quantity of a Product Step 2: turn moles into grams: – 0.63 mol Cu 2 S x (159g Cu 2 S/1 mol Cu 2 S) = g

10 Using a Limiting Reagent to Find the Quantity of a Product 25. How many grams of water can be produced by the reaction of 2.40 mol C 2 H 2 with 7.4 mol O 2 ?

11 Using a Limiting Reagent to Find the Quantity of a Product 26. If 2.70 mol C2H4 is reacted with 6.30 mol O2? – C 2 H 4 (g) + 2O 2 (g)  2CO(g) + 2H 2 O(g) A. identify the limiting reagent. B. Calculate the moles of water produced.

12 Percent Yield When an equation is used to calculate the amount of product that will form during a reaction, a value representing the theoretical yield is obtained. The theoretical yield is the max amount of product that could be formed from given amounts of reactants.

13 Percent Yield The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield. The actual yield is often less than the theoretical yield.

14 Percent Yield The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. Percent yield = (actual yield/theoretical yield) x 100%

15 Percent Yield In calculating the ratio of the actual yield to the theoretical yield, the percent yield is a measure of the efficiency of the reaction. A percent yield should not normally be larger than 100%, but can be lower due to reactions not going to completion.

16 Calculating the Theoretical Yield of a Reaction What is the theoretical yield of CaO if 24.8g CaCO 3 is heated? – CaCO 3 (s)  CaO(s) + CO 2 (g) Use the 3 Steps g CaCO3  mol CaCO3  mol CaO  g CaO

17 Calculating the Theoretical Yield of a Reaction 24.8g CaCO3 x (1 mol CaCO3/100.1g CaCO3) =.248 mol CaCO3.248 mol CaCO3 x (1 mol CaO/1 mol CaCO3) =.248 mol CaO.248 mol CaO x (56.1g CaO/1 mol CaO) = g CaO

18 Calculating the Theoretical Yield of a Reaction 27. When 84.8 g of iron(III) oxide reacts with an excess of carbon monoxide, iron is produced. What is the theoretical yield of this reaction? – Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)

19 Calculating the Theoretical Yield of a Reaction 28. When 5g of copper reacts with excess silver nitrate, silver metal and copper nitrate are produced. What is the theoretical yield of silver in this reaction?

20 Calculating the Percent Yield of a Reaction What is the percent yield of the reaction if the theoretical yeild of CaO is 13.9g and 13.1 g CaO is produced? % yield = (actual yield/theoretical yield) x 100% = (13.1g/13.9g) x 100% = 94.2%

21 Calculating the Percent Yield of a Reaction 29. if 50g of silicon dioxide is heated with an excess of carbon, 27.9g of silicon carbide is produced. What is the percent ield of this reaction?

22 Calculating the Percent Yield of a Reaction If 15g of nitrogen reacts with 15g of hydrogen, 10.5g of ammonia is produced. What is the percent yield of this reaction?


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