Chapter 9 Chemical Quantities.

Chapter 9 Chemical Quantities

9.1 Information Given by Chemical Equations
9.2 Mole–Mole Relationships 9.3 Mass Calculations 9.4 The Concept of Limiting Reactants 9.5 Calculations Involving a Limiting Reactant 9.6 Percent Yield Copyright © Cengage Learning. All rights reserved

A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction. The coefficients of a balanced equation give the relative numbers of molecules. Copyright © Cengage Learning. All rights reserved

C2H5OH + 3O2  2CO2 + 3H2O The equation is balanced.
All atoms present in the reactants are accounted for in the products. 1 molecule of ethanol reacts with 3 molecules of oxygen to produce 2 molecules of carbon dioxide and 3 molecules of water. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. Copyright © Cengage Learning. All rights reserved

Concept Check One of the key steps in producing pure copper from copper ores is the reaction below: Cu2S(s) + Cu2O(s) → Cu(s) + SO2(g) (unbalanced) After balancing the reaction, determine how many dozen copper atoms could be produced from a dozen molecules of the sulfide and two dozen of the oxide of copper(I). Also, how many moles of copper could be produced from one mole of the sulfide and two moles of the oxide? a) 1 dozen and 1 mole b) 2 dozen and 2 moles c) 3 dozen and 3 moles d) 6 dozen and 6 moles The correct answer is d. The balanced equation is: Cu2S(s) + 2Cu2O(s) → 6Cu(s) + SO2(g) Copyright © Cengage Learning. All rights reserved

2 mol of H2O yields 2 mol of H2 and 1 mol of O2.
A balanced equation can predict the moles of product that a given number of moles of reactants will yield. 2 mol of H2O yields 2 mol of H2 and 1 mol of O2. 4 mol of H2O yields 4 mol of H2 and 2 mol of O2. 2H2O(l) → 2H2(g) + O2(g) Copyright © Cengage Learning. All rights reserved

Consider the following balanced equation:
Example Consider the following balanced equation: Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s) How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na2SiF6? Where are we going? We want to determine the number of moles of NaF produced by Na with excess Na2SiF6. What do we know? The balanced equation. We start with 3.50 mol Na. Copyright © Cengage Learning. All rights reserved

Consider the following balanced equation:
Example Consider the following balanced equation: Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s) How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na2SiF6? How do we get there? Copyright © Cengage Learning. All rights reserved

Exercise Propane, C3H8, is a common fuel used in heating homes in rural areas. Predict how many moles of CO2 are formed when 3.74 moles of propane are burned in excess oxygen according to the equation: C3H8 + 5O2 → 3CO2 + 4H2O a) 11.2 moles b) 7.48 moles c) 3.74 moles d) 1.25 moles The correct answer is a. (3.74 mol C3H8) × (3 mol CO2/1 mol C3H8) = 11.2 mol CO2 Copyright © Cengage Learning. All rights reserved

Balance the equation for the reaction.
Steps for Calculating the Masses of Reactants and Products in Chemical Reactions Balance the equation for the reaction. Convert the masses of reactants or products to moles. Use the balanced equation to set up the appropriate mole ratio(s). Use the mole ratio(s) to calculate the number of moles of the desired reactant or product. Convert from moles back to masses. Copyright © Cengage Learning. All rights reserved

Stoichiometry Stoichiometry is the process of using a balanced chemical equation to determine the relative masses of reactants and products involved in a reaction. Copyright © Cengage Learning. All rights reserved

For the following unbalanced equation: Cr(s) + O2(g) → Cr2O3(s)
Example For the following unbalanced equation: Cr(s) + O2(g) → Cr2O3(s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? Where are we going? We want to determine the mass of Cr2O3 produced by Cr with excess O2. Copyright © Cengage Learning. All rights reserved

Example For the following unbalanced equation: Cr(s) + O2(g) → Cr2O3(s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? What do we know? The unbalanced equation. We start with 15.0 g Cr. We know the atomic masses of chromium (52.00 g/mol) and oxygen (16.00 g/mol) from the periodic table. Copyright © Cengage Learning. All rights reserved

Example For the following unbalanced equation: Cr(s) + O2(g) → Cr2O3(s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? What do we need to know? We need to know the balanced equation. 4Cr(s) + 3O2(g) → 2Cr2O3(s) We need the molar mass of Cr2O3. g/mol Copyright © Cengage Learning. All rights reserved

Convert the mass of Cr to moles of Cr.
How do we get there? 4Cr(s) + 3O2(g) → 2Cr2O3(s) Convert the mass of Cr to moles of Cr. Determine the moles of Cr2O3 produced by using the mole ratio from the balanced equation. Copyright © Cengage Learning. All rights reserved

Convert moles of Cr2O3 to grams of Cr2O3.
How do we get there? 4Cr(s) + 3O2(g) → 2Cr2O3(s) Convert moles of Cr2O3 to grams of Cr2O3. Conversion string: Copyright © Cengage Learning. All rights reserved

Reaction Stoichiometry
C6H H2  C6H12 3 Moles level How many grams of cyclohexane C6H12 can be made from 4.5 grams of benzene C6H6? Balanced? Use the table method. ? Grams level 4.5g ? 1 mole C6H6 1 mole C6H g C6H12 4.8 g C6H12 4.5 g C6H = g 1 mole C6H mole C6H12 How many grams of H2 will be needed? 1 mole C6H6 3 moles H g H2 .35 g H2 4.5 g C6H = g 1 mole C6H mole H2 Copyright © Cengage Learning. All rights reserved

C6H12O O2  CO H2O 6 6 6 How many grams of sugar are needed to make 6.23g water? ? 12.84 g ? 6.23 g 1 molew moles gs 6.23 gw = 10.4 gs 18.015gw 6 molew moles How many grams of carbon dioxide can be produced from g of oxygen? 1 moleo molec gc 12.84 go = 17.66 gc 31.998go 6 moleo 1 molec Copyright © Cengage Learning. All rights reserved

Exercise Tin(II) fluoride is added to some dental products to help prevent cavities. Tin(II) fluoride is made according to the following equation: Sn(s) + 2HF(aq)  SnF2(aq) + H2(g) How many grams of tin(II) fluoride can be made from 55.0 g of hydrogen fluoride if there is plenty of tin available to react? a) 431 g b) 215 g c) 72.6 g d) 1.37 g The correct answer is b. (55.0 g HF) × (1 mol HF/ g HF) × (1 mol SnF2/2 mol HF) × ( g SnF2/1 mol SnF2) = 215 g SnF2 Copyright © Cengage Learning. All rights reserved

Exercise Consider the following reaction: PCl3 + 3H2O  H3PO3 + 3HCl
What mass of H2O is needed to completely react with 20.0 g of PCl3? 7.87 g 0.875 g 5.24 g 2.62 g The correct answer is a. (20.0 g PCl3) × (1 mol PCl3/ g PCl3) × (3 mol H2O/1 mol PCl3) × ( g H2O/1 mol H2O) = 7.87 g H2O Copyright © Cengage Learning. All rights reserved

Exercise Consider the following reaction where X represents an unknown element: 6 X(s) B2O3(s)  B4X3(s) XO2(g) If 175 g of X reacts with diboron trioxide to produce 2.43 moles of B4X3, what is the identity of X? a) Ge b) Mg c) Si d) C The correct answer is d. The identity of X is carbon. (2.43 mol B4X3) × (6 mol X/1 mol B4X3) = 14.6 mol X Molar Mass = # grams/# moles = 175 g X/14.6 mol X = 12.0 g/mol Copyright © Cengage Learning. All rights reserved

Stoichiometric Mixture
Contains the relative amounts of reactants that matches the numbers in the balanced equation. N2(g) + 3H2(g)  2NH3(g) Copyright © Cengage Learning. All rights reserved

Limiting Reactant Mixture
N2(g) + 3H2(g)  2NH3(g) Copyright © Cengage Learning. All rights reserved

Limiting Reactant Mixture
N2(g) + 3H2(g)  2NH3(g) Limiting reactant is the reactant that runs out first and thus limits the amounts of product(s) that can form. H2 Copyright © Cengage Learning. All rights reserved

H2O molecules are used up first, leaving two CH4 molecules unreacted.
Limiting Reactants H2O molecules are used up first, leaving two CH4 molecules unreacted. Copyright © Cengage Learning. All rights reserved

The amount of products that can form is limited by the water.
Limiting Reactants The amount of products that can form is limited by the water. Water is the limiting reactant. Methane is in excess. Copyright © Cengage Learning. All rights reserved

Steps for Solving Stoichiometry Problems Involving Limiting Reactants
1. Write and balance the equation for the reaction. 2. Convert known masses of reactants to moles. 3. Using the numbers of moles of reactants and the appropriate mole ratios, determine which reactant is limiting. 4. Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. 5. Convert from moles of product to grams of product, using the molar mass (if this is required by the problem). Copyright © Cengage Learning. All rights reserved

Example You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to know in order to determine the mass of product that will be produced? Copyright © Cengage Learning. All rights reserved

Where are we going? What do we need to know?
Let’s Think About It Where are we going? To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B. What do we need to know? The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation. The molar masses of A, B, and the product they form. Copyright © Cengage Learning. All rights reserved

Example – Continued You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C Copyright © Cengage Learning. All rights reserved

Convert known masses of reactants to moles.
Example – Continued You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C How do we get there? Convert known masses of reactants to moles. Copyright © Cengage Learning. All rights reserved

Determine which reactant is limiting.
Example – Continued You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C How do we get there? Determine which reactant is limiting. react with all of the A Only mol B is available, so B is the limiting reactant. Copyright © Cengage Learning. All rights reserved

Compute the number of moles of C produced.
Example – Continued You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C How do we get there? Compute the number of moles of C produced. Copyright © Cengage Learning. All rights reserved

Convert from moles of C to grams of C using the molar mass.
Example – Continued You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C How do we get there? Convert from moles of C to grams of C using the molar mass. Copyright © Cengage Learning. All rights reserved

Notice We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. Copyright © Cengage Learning. All rights reserved

2 H I2  HI How many grams of HI can be formed from 2.00 g H2 and 2.00 g of I2? 2.00 g 2.00 g ? 1 mole H2 2 mole HI g HI 2.00 g H = 254 g HI g 1 mole H mole HI 1 mole I2 2 moles HI g HI 2.00g I = 2.02 g HI g 1 mole I mole HI Limiting Reactant is always the smallest value! I2 is the limiting reactant and H2 is in XS. Copyright © Cengage Learning. All rights reserved

5 3 4 C3H O2  CO H2O How many grams of CO2 can be formed from 1.44 g C3H8 and 2.65 g of O2? 1.44 g 2.65g ? 1 mole C3H8 3 mole CO g CO2 4.31 g CO2 1.44 g C3H = g mole C3H mole CO2 1 mole O2 3 moles CO g CO2 2.19 g CO2 2.65g O = g mole O mole CO2 O2 is the limiting reactant and C3H8 is in XS. Copyright © Cengage Learning. All rights reserved

a) 2 moles of H2 and 2 moles of O2 2 moles of H2 and 3 moles of O2
Concept Check Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H2 + O2  2H2O a) 2 moles of H2 and 2 moles of O2 2 moles of H2 and 3 moles of O2 2 moles of H2 and 1 mole of O2 d) 3 moles of H2 and 1 mole of O2 e) Each produce the same amount of product. The correct answer is e. All of these reaction mixtures would produce two moles of water. Copyright © Cengage Learning. All rights reserved

Concept Check Consider the equation: A + 3B  4C. If 3.0 moles of A is reacted with 6.0 moles of B, which of the following is true after the reaction is complete? a) A is the leftover reactant because you only need 2 moles of A and have 3. b) A is the leftover reactant because for every 1 mole of A, 4 moles of C are produced. c) B is the leftover reactant because you have more moles of B than A. d) B is the leftover reactant because 3 moles of B react with every 1 mole of A. The correct answer is a. (6.0 mol B) × (1 mol A/3 mol B) = 2.0 mol A needed Copyright © Cengage Learning. All rights reserved

Exercise How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process? N2 + 3H2 → 2NH3 a) 2.00 g b) 3.00 g c) 3.64 g d) 1.82 g 3.00 g N2(1mole/28g N2)(2 moles NH3/1 moles N2) (17 g NH3/1 mole NH3) = g NH3 The correct answer is c. Nitrogen is the limiting reactant. (0.107 mol N2) × (2 mol NH3/1 mol N2) × ( g NH3/1 mol NH3) = 3.65 g NH3 3.00 g H2(1mole/2g H2)(2 moles NH3/3 moles H2) (17 g NH3/1 mole NH3) = g NH3 Copyright © Cengage Learning. All rights reserved

Theoretical Yield The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. The actual yield (amount actually produced) of a reaction is usually less than the maximum expected (theoretical yield). Copyright © Cengage Learning. All rights reserved

We determined that 8.33 g C should be produced.
Example – Recall You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C We determined that 8.33 g C should be produced. 7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction? Where are we going? We want to determine the percent yield of the reaction. Copyright © Cengage Learning. All rights reserved

We know the actual and theoretical yields.
Example – Recall You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C We determined that 8.33 g C should be produced g of C is what was actually made in the lab. What is the percent yield of the reaction? What do we know? We know the actual and theoretical yields. Copyright © Cengage Learning. All rights reserved

Example – Recall You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C We determined that 8.33 g C should be produced g of C is what was actually made in the lab. What is the percent yield of the reaction? How do we get there? Copyright © Cengage Learning. All rights reserved

Exercise Find the percent yield of product if 1.50 g of SO3 is produced from 1.00 g of O2 and excess sulfur via the reaction: 2S + 3O2 → 2SO3 a) 40.0% b) 80.0% c) 89.8% d) 92.4% First we find the maximum yield (theoretical yield) of product: (1.00 g O2)(1 mol O2/32.00 g O2)(2 mol SO3/3 mol O2)(80.06 g SO3/1 mol SO3) = 1.67 g (theoretical yield) We then divide the actual yield by the theoretical yield: Percent yield = (1.50 g/1.67 g) × 100 = 89.8% Copyright © Cengage Learning. All rights reserved

Consider the following reaction: P4(s) + 6F2(g)  4PF3(g)
Exercise Consider the following reaction: P4(s) + 6F2(g)  4PF3(g) What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield? a) 29.9 g b) 46.1 g c) 184 g d) 738 g 64.9% = (85.0 g PF3 / x)(100); x = g PF3 ( g PF3)(1 mol PF3 / g PF3)(1 mol P4 / 4 mol PF3)( g P4 / 1 mol P4) = 46.1 g of P4 needed Copyright © Cengage Learning. All rights reserved

Assignment: Read 5.6 up to sample (238-9)
Define the following terms: yield, theoretical yield, actual yield, percentage yield. Based on your reading, give 4 reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. Read the sample problem on the next slide and try the practice problem on slide number 5 When 5.00 g of KClO3 is heated it decomposes according to the equation: 2KClO3  2KCl + 3O2 a) Calculate the theoretical yield of oxygen. b) Give the % yield if 1.78 g of O2 is produced. c) How much O2 would be produced if the percentage yield was 78.5%? Copyright © Cengage Learning. All rights reserved

Yield: the amount of product
Answers 1) Yield: the amount of product Theoretical yield: the amount of product we expect, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is given in the question) Percent yield: = x 100% actual yield theoretical yield 2) Not all product is recovered (e.g. spattering) Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk) A side reaction occurs (e.g. MgO vs. Mg3N2) The reaction does not go to completion Copyright © Cengage Learning. All rights reserved

Step 3: Calculate % yield
Sample problem Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2  2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 16 g H2 143 g = Step 3: Calculate % yield actual theoretical 138 g H2O 143 g H2O = % yield = x 100% x 100% 96.7% = Copyright © Cengage Learning. All rights reserved

Step 3: Calculate % yield
Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2  2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH3 3 mol H2 x 17.04 g NH3 1 mol NH3 x # g NH3= 20.0 mol H2 227 g = Step 3: Calculate % yield actual theoretical 40.5 g NH3 227 g NH3 = % yield = x 100% x 100% 17.8% = Copyright © Cengage Learning. All rights reserved

# g O2= (also works if you use mol O2)
Answers 4) 2KClO3  2KCl + 3O2 a) b) c) # g O2= (also works if you use mol O2) 1 mol KClO3 g KClO3 x 3 mol O2 2 mol KClO3 x 32 g O2 1 mol O2 x 5.00 g KClO3 1.958 g = actual theoretical 1.78 g O2 1.958 g O2 = % yield = x 100% x 100% 90.9% = actual theoretical x g O2 1.958 g O2 = % yield = x 100% x 100% 78.5% = 78.5% x g O2 100% = x g O2 1.537 g O2 = Copyright © Cengage Learning. All rights reserved

Hint: determine limiting reagent first
Challenging question 2H2 + O2  2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first 1 mol O2 32 g O2 x 2 mol H2O 1 mol O2 x 18.02 g H2O 1 mol H2O x # g H2O= 60 g O2 68 g = 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 7.0 g H2 62.4 g = actual theoretical 58 g H2O 62.4 g H2O = % yield = x 100% x 100% 92.9% = Copyright © Cengage Learning. All rights reserved

More Percent Yield Questions
Note: try “shortcut” for limiting reagent problems The electrolysis of water forms H2 and O H2O  2H2 + O What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2 What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS Copyright © Cengage Learning. All rights reserved

More Percent Yield Questions
Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 Copyright © Cengage Learning. All rights reserved

The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2
Q1 The electrolysis of water forms H2 & O H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? Actual yield is given: 12.3 g O2 Next, calculate theoretical yield # g O2= 1 mol H2O 18.02 g H2O x 1 mol O2 2 mol H2O x 32 g O2 1 mol O2 x 14.0 g H2O 12.43 g = Finally, calculate % yield actual theoretical 12.3 g O2 12.43 g O2 = % yield = x 100% x 100% 98.9% = Copyright © Cengage Learning. All rights reserved

Actual yield is given: 107 g O2 Next, calculate theoretical yield
Q2 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3  2KCI + 3O2 Actual yield is given: 107 g O2 Next, calculate theoretical yield # g O2= 1 mol KClO3 g KClO3 x 3 mol O2 2 mol KClO3 x 32 g O2 1 mol O2 x 300 g KClO3 117.5 g = Finally, calculate % yield actual theoretical 107 g O2 117.5 g O2 = % yield = x 100% x 100% 91.1% = Copyright © Cengage Learning. All rights reserved

Actual yield is given: 220 g FeS Next, calculate theoretical yield
Q3 What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS Actual yield is given: 220 g FeS Next, calculate theoretical yield 1 mol FeS 1 mol Fe x 87.91 g FeS 1 mol FeS x # g FeS= 3.00 mol Fe 263.7 g = Finally, calculate % yield actual theoretical 220 g O2 263.7 g O2 = % yield = x 100% x 100% 83.4% = Copyright © Cengage Learning. All rights reserved

# g Fe2O3= 1 mol FeS2 119.97 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x
4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 1 mol FeS2 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x 159.7 g Fe2O3 1 mol Fe2O3 x 300 g FeS2 199.7 g Fe2O3 = 1 mol O2 32 g O2 x 2 mol Fe2O3 11 mol O2 x 159.7 g Fe2O3 1 mol Fe2O3 x 200 g O2 g Fe2O3 = actual theoretical 143 g Fe2O3 g Fe2O3 = % yield = x 100% x 100% 78.8% = Copyright © Cengage Learning. All rights reserved

For more lessons, visit www.chalkbored.com
70 g of MnO mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 # g Cl2= 1 mol MnO2 86.94 g MnO2 x 1 mol Cl2 1 mol MnO2 x 70.9 g Cl2 1 mol Cl2 x 70 g MnO2 57.08 g Cl2 = 1 mol Cl2 4 mol HCl x 71 g Cl2 1 mol Cl2 x 3.5 mol HCl 62.13 g Cl2 = actual theoretical x g Cl2 57.08 g Cl2 = % yield = x 100% x 100% 42% = 42% x g Cl2 100% = x g Cl2 24 g Cl2 = For more lessons, visit Copyright © Cengage Learning. All rights reserved

Chapter 9 Chemical Quantities.

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