Advanced Stoichiometry - PowerPoint Presentation

Advanced Stoichiometry Chapter 9, Section 2 (pages 312 – 318) Problem Set E (p. 314 # 1 – 3) Problem Set F (p. 317 # 1 – 3)

Terms to Know and Understand: Limiting Reactant – the substance that controls the quantity of product that can form in a chemical reaction. (May not be the reactant having the lowest mass!) Identified through stoichiometry . Excess Reactant – the substance that is not used up completely in a reaction. Theoretical Yield – the maximum quantity of product that a reaction could theoretically make if everything about the reaction works perfectly. The theoretical yield is found by stoichiometry . Actual Yield – the quantity of product actually produced in a reaction. The actual yield is found experimentally. Usually less than the theoretical yield. Percentage Yield – the ratio relating the actual yield of a reaction to its theoretical yield.

Identifying a limiting reactant: Example: A chemical reaction occurs between copper (II) oxide and hydrogen according to the following balanced equation: CuO (s) + H 2 (g)  Cu (s) + H 2 O (g) What is the limiting reactant when 19.9-g of CuO react with 2.02-g H 2 ? What is the theoretical yield of copper? First read through other questions in the problem to see if you will be asked to find the theoretical yield of one of the products – if this is the case, do stoichiometry using this product as your ‘find’. If you are not asked to find the mass of one of the products, choose one as your ‘find’. Do the stoichiometry twice using each reactant as your ‘given’. To do the remaining work I will use Cu (s) as my find’.

Identifying a limiting reactant: Using CuO as the given: 19.9 g CuO 1 mol CuO 79.545 g CuO 0.250 mol CuO 0.250 mol Cu 1 mol Cu 1 mol CuO 0.250 mol CuO 15.9 g Cu 63.546 g Cu 1 mol Cu 0.250 mol Cu Mass of copper that would be produced if CuO was the limiting reactant. Step 1 Step 2 Step 3

Identifying a limiting reactant: Using H 2 as the given: 2.02 g H 2 1 mol H 2 2.0158 g H 2 1.00 mol H 2 1.00 mol Cu 1 mol Cu 1 mol H 2 1.00 mol H2 63.5 g Cu 63.546 g Cu 1 mol Cu 1.00 mol Cu Mass of copper that would be produced if H 2 was the limiting reactant. Step 1 Step 2 Step 3

Compare the calculated quantities of product: When 15.9-g of Cu are produced, the 19.9-g of CuO will be completely consumed (used up). If 63.5-g of Cu were produced, 2.02 g of H 2 would be completely consumed, but there is not enough CuO to do this.The limiting reactant is CuO, because it will be gone after 15.9-g of Cu is produced, and the reaction will stop.The excess reactant is H2, because the amount available would allow the reaction to produce 63.5-g Cu, but the reaction stops when the limiting reactant is used up.The theoretical yield of Cu will be 15.9-g, because the reaction will stop as soon as this amount of product is formed.Identifying a limiting reactant:

Note that CuO is the limiting reactant despite the fact that there was a lesser mass of H 2 available to react – it is the stoichiometric quantity that is important! If you are only asked to find the theoretical yield of a product you must still identify the limiting reactant, as it will determine the amount of product formed!You can recognize limiting reactant problems because you are given amounts of two of the reactants in the problem.Identifying a limiting reactant:

Finding the leftover amount of excess reactant: In our example problem, we note that we had 19.9-g of CuO and 2.02-g H 2 available to react. Since the CuO is the limiting reactant, it will be completely consumed.Since H2 is the excess reactant, some will be left over.We can find the amount of H2 leftover if we subtract the amount of H2 consumed (reacted) from the amount of H2 available.We will use stoichiometry to find the amount of H2 consumed during the reaction. For all problems involving a limiting reactant, your ‘given’ is the mass of limiting reactant available to you.

Finding the leftover amount of excess reactant: Using CuO ( the limiting reactant ) as the given : 19.9 g CuO 1 mol CuO 79.545 g CuO 0.250 mol CuO 0.250 mol H 2 1 mol H 2 1 mol CuO0.250 mol CuO 0.504 g H 2 2.0158 g H 2 1 mol H2 0.250 mol H2 Mass of H 2 that will be consumed during the reaction. Step 1 Step 2 Step 3

Finding the leftover amount of excess reactant: Calculate the mass of H 2 (excess reactant) leftover: Mass of excess reactant available (from problem): Subtract mass of excess reactant consumed (from stoichiometry): Mass of excess reactant leftover:2.02-g H2– 0.504-g H2 1.52-g H2

Determining Percentage Yield percentage yield = actual yield theoretical yield x 100 For example, say we conduct the reaction between CuO and H 2 with the amounts given in the example problem on slide 3. While we expect to produce 15.9-g of Cu (the theoretical yield), we find that we are only able to collect 13.8-g of Cu following the reaction. We can calculate the percentage yield as follows:percentage yield = 13.8-g 15.9-g x 100 percentage yield = 86.8%

What mass of silver sulfide, Ag 2 S, can be made from 123g of H 2 S obtained from a rotten egg? 4Ag + 2H 2S +O2  2Ag2S + 2H2O

What mass, in grams, of water is produced when 80 liters of hydrogen gas react with oxygen? 2H 2 + O 2  2H 2 O

•       Practice Box E (p. 314 # 1 – 3) •       Chapter 9 Review, p. 331 # 31, 32, 33

Determining Percentage Yield A typical problem will give you the actual yield , but require that you first identify the limiting reactant, and use stoichiometry to calculate the theoretical yield: Calculate the percentage yield of H 3 PO4 if 126.2-g are recovered when 100.0-g of P4O10 react with 200.0-g H2O according to the following balanced equation:P4O10 + 6 H2O  4 H3PO4Identify the information given to us by the problem:actual yield of H3PO4 is 126.2-gavailable amounts of reactants:100.0-g of P 4O10200.0-g H 2ODetermine what we need to calculate first the limiting reactantthe theoretical yield of H3PO 4

Determining Percentage Yield – Find the Limiting Reactant and Theoretical Yield Using P 4 O 10 as the given: Using H 2 O as the given: 100.0 g P 4 O 101 mol P4O10283.882 g P4O100.3522 mol P4O10 1.409 mol H 3 PO 44 mol H3PO41 mol P4 O10 0.3522 mol P4O10 138.1 g H 3 PO 4 97.9927 g H 3 PO 4 1 mol H 3 PO 4 1.409 mol H 3 PO 4 Step 1 Step 2 Step 3 O 200.0 g H 2 O 1 mol H 2 O 18.0148 g H 2 O 11.10 mol H 2 O 7.400 mol H 3 PO 4 4 mol H 3 PO 4 6 mol H 2 O 11.10 mol H 2 O 725.1 g H 3 PO 4 97.9927 g H 3 PO 4 1 mol H 3 PO 4 7.400 mol H 3 PO 4 Step 1 Step 2 Step 3 theoretical yield limiting reactant

Determining Percentage Yield percentage yield = actual yield theoretical yield x 100 percentage yield = 126.2 g 138.1 g x 100 percentage yield = 91.38%

Another type of percentage yield problem: Calculating Actual Yield In these problems you are given the percentage yield and either: the theoretical yield, or amounts of each reactant determine limiting reactant calculate theoretical yield Example: How many grams of PbI 2 should form if 355-g are theoretically possible and the percentage yield for the reaction is 84.5%?

Calculating Actual Yield percentage yield = actual yield theoretical yield x 100 rearrange the equation: percentage yield = actual yield theoretical yield x 100 /100 /100 actual yieldtheoretical yieldpercentage yield100= x355 g 84.5 100 =x0.845 (355 g)=x (actual yield) 300. g =