Bell Ringer 4/4/16 When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield? SO3 + H2O H2SO4 1 mol SO3 1 mol H2SO4 98.09 g H2SO4 6.58 g SO3 x x x = 80.07 g SO3 1 mol SO3 1 mol H2SO4 8.06 g H2SO4 1 mol H2O 1 mol H2SO4 98.09 g H2SO4 1.64 g H2O x x x = 18.02 g H2O 1 mol H2O 1 mol H2SO4 8.93 g H2SO4

Bell Ringer When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is acid, what is the percent yield? SO3 + H2O H2SO4 Info we’ve learned: Theoretical Yield = 8.06 g H2SO4 7.99 g H2SO4 % Yield = Actual Yield Theoretical Yield x 100 % = 99.1 % 8.06 g H2SO4

Selected answers Homework p.231#25-40/p.233 #44-50 25. 2KClO3(s) 2KCl + 3O2(g) Two formula units of 2KClO3 decompose to make two formula units of KCl and three molecules of O2. 26. 2KClO3(s) 2KCl + 3O2(g) Two moles of 2KClO3 decompose to make two moles of KCl and three molecules of O2. 27. a. 2KClO3(s) 2KCl + 3O2(g) 2[39.1+35.45 + 48.0 ] = 245.1g = 2[39.1 + 35.45] + 3[32.0] 245.1g = 149.1 + 64.0 = 245.1g b.248.0g c. 188.4g d. All obey law 28.a.0.54 mol b. 13.6 mol c.0.984 mol d. 236 mol

29. _______g 4.80g 2 Na2O2 (s) + 2 H2O(l)  O2 (g) + 4 NaOH(aq) _ 0.30_moles __0.15___moles 4.80 g O2 1 mole O2 2 moles Na2O2 78.0g Na2O2 ---------------------------------------------------------------------------------------------------------- = 32.0 g O2 1 mole O2 1 moles Na2O2 Gram to mole step mole ratio step mole to grams Answer: 29. a 23.4 grams

29.b 24.0g NaOH c. 0.098 g O2 30. Paper vs. plastic? 31. a. 11.3 mol CO b. 112 g CO, 16.0 g H2 c. 11.4 g H2 32.a. 372 g F2 b. 1.32 g NH3 c. 123 g N2F4 33.a. 51.2 g H2O b. 5.71 X 1023 molecules NH3 c. 23.2 g Li2N 34. Choose one given and solve for the other. If I need more than I was given it is the L.R. because it will be used up to complete the reaction. 35. a. Cl2 b. H2 c. H2O d. O2 36. a. 3.5 mol AlCl3 b. 6.4 mol H2O c. 1.01 mol H3PO4 b. 3.6 mol P4O10 37. a. 1 mol Al 38. b. 0.2 mol O2 c. 0.23mol P4O10 d. 0.1 mol P 38. 43.2 g H2O

40. The “efficiency” of a reaction 10 mole Cmp A 4 mol Cmp B 2 mol Cmp C ?= 0.5 mol Cmp D ?= 0.0500 mol Cmp E 0.0100 mol Cmp F 39. 40. The “efficiency” of a reaction % = 100 (actual/theoretical )= percent yield 40% yield 50 % yield 25% 10% 20%

44. Given: 3.74 g Ca3(PO4)2 Find: grams H3PO4 grams /molar mass = moles Ratio: 1 mol Ca3(PO4)2 = 2 mole H3PO4 3. moles H3PO4 * molar mass = grams 47. a. 700 L N2 b. no reagent in excess 48.a. 96.4% b. 46.6 g CO a. 2.36 g H3PO4 1.28 L CO2 5.70 X 1021 atoms of Zn b. 32.7 mg Zn 10.7 kg CaSO4 a. 31.92 L b. 40.2 g H2O 46.a. 10.7 g NO, 12.8 g H2O b. 95.2 g Zn

Stoichiometry Review

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Mole-Mole Problems 2 H2O 2 H2 + O2 1 conversion step Given: moles “A” Required: moles “B” Convert moles “A” to moles “B” using mole ratio. The mole ratio is used in EVERY STOICHIOMETRY PROBLEM. EVER. I PROMISE. 2 H2 + O2 2 H2O How many moles of water can be formed from 0.5 mol H2? 2 mol H2O 0.5 mol H2 0.5 mol H2O x = 2 mol H2

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Mass-Mole Problems 2 H2O 2 H2 + O2 2 conversion steps Given: mass “A” Required: moles “B” Step 1: convert grams “A” to moles “A” using Periodic Table Step 2: convert moles “A” to moles “B” using mole ratio 2 H2 + O2 2 H2O How many moles of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 48.0 g O2 3.00 mol H2O x x = 32.00 g O2 1 mol O2

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Mass-Mass Problems 2 H2O 2 H2 + O2 3 conversion steps Given: mass “A” Required: mass “B” Step 1: convert grams “A” to moles “A” using Periodic Table Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to grams “B” using Periodic Table 2 H2 + O2 2 H2O How many grams of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 18.02 g H2O 48.0 g O2 54.1 g H2O x x x = 32.00 g O2 1 mol O2 1 mol H2O

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Mass-Volume Problems 2 H2O 2 H2 + O2 3 – 4 conversion steps Given: mass “A” Required: volume “B” Step 1: convert grams “A” to moles “A” using Periodic Table Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to liters “B” 2 H2 + O2 2 H2O How many liters of oxygen are necessary to create 48.0 g H2O? 48.0 g H2O 1 mol H2O 1 mol O2 22.4 L O2 29.8 L O2 x x x = 18.02 g H2O 2 mol H2O 1 mol O2

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

How many grams of water are formed by reacting 36.0 L O2? Volume-Mass Problems 3 – 4 conversion steps Given: volume “A” Required: mass “B” Step 1: convert liters “A” to moles “A” Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to grams “B” using Periodic Table 2 H2 + O2 2 H2O How many grams of water are formed by reacting 36.0 L O2? 36.0 L O2 1 mol O2 2 mol H2O 18.02 g H2O 58.7 g H2O = x x x 22.4L O2 1 mol O2 1 mol H2O

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Volume-Volume Problems 3 – 5 conversion steps Given: volume “A” Required: volume “B” Step 1: convert liters “A” to moles “A” Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to liters “B” 2 H2 + O2 2 H2O How many liters of H2 are required to react with 5.0 L O2? 5.0 L O2 1 mol O2 2 mol H2 22.4 L H2 10. L H2 x x x = 22.4 L O2 1 mol O2 1 mol H2

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Limiting Reactant Problems Quantities are given for each reactant. 2 parallel equations Solve each equation for product desired and determine limiting reactant. Use Limiting Reactant to solve for amount or excess reactant used. Subtract amount excess reactant used from amount given to determine how much is left over.

Limiting Reactant Problems 2 H2 + O2 2 H2O If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be? 1 mol O2 2 mol H2O 18.02 g H2O 10.0 g O2 x x x = 32.00 g O2 1 mol O2 1 mol H2O LIMITING REACTANT 11.3 g H2O THEORETICAL YIELD 5.00 g H2 1 mol H2 2 mol H2O 18.02 g H2O x x x = EXCESS REACTANT 2.02 g H2 2 mol H2 1 mol H2O 44.06 g H2O

Limiting Reactant Problems 2 H2 + O2 2 H2O If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be? Info we know so far: Limiting Reactant = O2 Excess Reactant = H2 1 mol O2 2 mol H2 2.02 g H2 10.0 g O2 x x = x 32.00 g O2 1 mol O2 1 mol H2 1.26 g H2 USED 5.00 g H2 – 1.26 g H2 = 3.74 g H2 LEFT OVER

Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

Percent Yield Problems Critical Information: Theoretical Yield Actual Yield Percent Yield May or may not be given You will be given one of these Determine the actual yield of a reaction between 6.25 g H2 and excess O2 that has a 85% percent yield. 2 H2 + O2 2 H2O 6.25 g H2 1 mol H2 2 mol H2O 18.02 g H2O x x x = 2.02 g H2 2 mol H2 1 mol H2O THEORETICAL YIELD 55.6 g H2O ? 85 % = 100 % x 55.6 g H2O ACTUAL YIELD = 47.3 g H2O