## Presentation on theme: "Copyright © Houghton Mifflin Company. All rights reserved. 9 | 1 Information Given by the Chemical Equation Balanced equations show the relationship between."— Presentation transcript:

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 1 Information Given by the Chemical Equation Balanced equations show the relationship between the relative numbers of reacting molecules and product molecules. 2 CO + O 2  2 CO 2 2 CO molecules react with 1 O 2 molecule to produce 2 CO 2 molecules.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 2 The information given is relative: 2 CO + O 2  2 CO 2 Therefore: 200 CO molecules react with 100 O 2 molecules to produce 200 CO 2 molecules. 2 moles CO molecules react with 1 mole O 2 molecules to produce 2 moles CO 2 molecules. 12 moles CO molecules react with 6 moles O 2 molecules to produce 12 moles CO 2 molecules. Information Given by the Chemical Equation

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 3 The coefficients in the balanced chemical equation also show the mole ratios of the reactants and products. Since moles can be converted to masses, we can also determine the mass ratios of the reactants and products. Information Given by the Chemical Equation

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 4 2 CO + O 2  2 CO 2 2 moles CO; 1 mole O 2 ; 2 moles CO 2 Since 1 mole of CO = 28.01 g (12.01g + 16g), 1 mole O 2 = 32.00 g (2 x 16.00 g), and 1 mole CO 2 = 44.01 g (12.01 g + 32 g), then we have the following quantities: 2(28.01) g CO; 1(32.00) g O 2 ; 2(44.01) g CO 2 Information Given by the Chemical Equation

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 5 Example #1 Determine the number of moles of carbon monoxide required to react with 3.2 moles oxygen, and determine the moles of carbon dioxide produced.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 6 Example #1 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 8 Example #2 Consider the following reaction: CH 4 (g) + 4Cl 2 (g)  CCl 4 (l) + 4HCl(g) How many moles of Cl 2 would be needed to react with 0.3 mol of CH 4 ?

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 9 Example #3 Determine the number of grams of carbon monoxide required to react with 48 g of oxygen, and determine the number of grams of carbon dioxide produced.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 10 Example #3 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationships: Determine the molar mass of each:

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 11 Use the molar mass of the given quantity to convert it to moles. Use the mole relationships to convert the moles of the given quantity to the moles of the desired quantity: Example #3 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 12 Use the molar mass of the desired quantity to convert the moles to mass: Example #3 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 13 Example #4 Consider the reaction below: Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) If you start with 0.056 mol of Zn, how much HCl in grams is needed? How much ZnCl 2 in grams can be produced?

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 14 Limiting and Excess Reactants Limiting reactant: a reactant that is completely consumed when a reaction is run to completion Excess reactant: a reactant that is not completely consumed in a reaction Theoretical yield: the maximum amount of a product that can be made by the time the limiting reactant is completely consumed

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 16 Example #5 Determine the number of moles of potassium oxide produced when 8 moles of potassium metal reacts with 3 moles of oxygen.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 18 Use dimensional analysis to determine the # of moles of reactant A needed to react with reactant B: Example #5 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 19 Compare the calculated # of moles of reactant A to the # of moles of reactant A given. If the calculated # of moles is greater, then A is the limiting reactant; if the calculated # of moles is less, then A is the excess reactant. What is the limiting reagent in this reaction? Example #5 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 20 Use the limiting reactant to determine the maximum # of moles of product that can form: Example #5 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 21 Example #6 Consider the reaction between zinc metal and iodine (I 2 ) to form zinc(II) iodide. If you begin with 50g of Zn and 50 g of I 2, what is the limiting reagent? How much zinc(II) iodide in grams can be formed in this reaction?

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 22 Example #7 Consider the following unbalanced equation: SF 4 + I 2 O 5  IF 5 + SO 2 Calculate the maximum number of grams of IF 5 that can be produced from 10 g of SF 4 and 10 g of I 2 O 5.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 23 Percent Yield Most reactions do not go to completion. Theoretical yield: The maximum amount of product that can be made in a reaction Actual yield: the amount of product actually made in a reaction Percent Yield = Actual Yield Theoretical Yield x 100%

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 24 Example #8 If in the last problem 9.0 g of IF 5 is actually made, what is the percent yield of IF 5 ?

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 25 Example #9 Consider the following reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g) When 10g of N 2 and 10 g of NH 3 react, 8 g of NH 3 is formed. What is the percent yield of NH 3 ?

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 26 Example #10 Consider the following reaction: CS 2 (l) + 3O 2 (g)  CO 2 (g) + 2SO 2 (g) When 1 g of CS 2 reacts with 1 g of O 2, 0.33 g of CO 2 results. What is the percent yield of CO 2 ?