Limiting Reagent and Percent Yield

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Limiting Reagent and Percent Yield

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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 reaction occurs only until the limiting reagent is used up. Limiting Agent. - PowerPoint PPT Presentation

Transcript of Limiting Reagent and Percent Yield

Page 1: Limiting Reagent and Percent Yield

Limiting Reagent and Percent Yield

Page 2: 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 reaction occurs only until the limiting reagent is used up.

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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.

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Determining the Limiting Reagent in a Reaction

• What is the limiting reagent when 80g Cu reacts with 25g S?– 2Cu(s) + S(s) Cu2S(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) = 0.779 mol S

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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) = 0.630 S

• Since the amount of sulfur needed is 0.630 moles and the given amount is 0.779 moles, this indicates that sulfur is in excess. Thus copper is the limiting reagent.

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Determining the Limiting Reagent in a Reaction - Practice

• 23. If 2.70 mol C2H4 is reacted with 6.30 mol O2, identify the limiting reagent.

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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) MgCl2(aq) + H2(g)

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Using a Limiting Reagent to Find the Quantity of a Product

• What is the maximum number of grams of Cu2S that can be formed from:– 2Cu(s) + S(s) Cu2S(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 Cu2S/2 mol Cu) = 0.63 mol Cu2S

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Using a Limiting Reagent to Find the Quantity of a Product

• Step 2: turn moles into grams:– 0.63 mol Cu2S x (159g Cu2S/1 mol Cu2S) = 100.17 g

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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 C2H2 with 7.4 mol O2?

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Using a Limiting Reagent to Find the Quantity of a Product

• 26. If 2.70 mol C2H4 is reacted with 6.30 mol O2?– C2H4(g) + 2O2(g) 2CO(g) + 2H2O(g)

• A. identify the limiting reagent.• B. Calculate the moles of water produced.

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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.

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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.

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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%

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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.

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Calculating the Theoretical Yield of a Reaction

• What is the theoretical yield of CaO if 24.8g CaCO3 is heated?– CaCO3(s) CaO(s) + CO2(g)

• Use the 3 Steps• g CaCO3 mol CaCO3 mol CaO g CaO

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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)• = 13.91 g CaO

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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)

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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?

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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%

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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?

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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?