Unit 14 Calculations for Chemical Equations INTRODUCTION The most often encountered problem in chemistry is one that involves a balanced chemical reaction. Almost all areas of chemistry deal with reactions of one type or another. The previous unit prepared you to interpret chemical equations. This unit will prepare you to work quantitatively with reactions using a consistent format. The format will allow you to work quite simple ones, or seemingly complex problems. They are all based on the principles of chemical reactions as well as some basic calculations. OBJECTIVES 1. The student will solve problems involving chemical equations, using the method presented in this unit to within the accuracy of the data provided. DISCUSSION A. Initial Considerations As mentioned in the previous unit all equations will be interpreted in terms of the number of moles of reactants and products. The numbers given by the coefficients in the balanced reaction provide us with conversion factors between products or more likely between a reactant and a product. These conversion factors are used just as any other conversion factor is used. They are simply the ratios between the number of moles of reactants and products. Example (1) In the reaction: H +0 H 0 The coefficients, 1 and can be made into the following conversion factors: mole H = 1 mole O Two moles of H always reacts with one mole of 0 mole H = mole H 0 Two moles of H always produce two moles of H 0 1 mole 0 = mole H 0 One mole of 0 always produces two moles of H 0 This is not to say that two moles of Hydrogen are exactly the same as two moles of Oxygen but that in this particular reaction this is always the mole ratio that they react by. These conversion factors are simply taken from the balanced chemical equation. The particular problem determines the combination.
B. Problem Solving Format In order to solve problems involving chemical equations you must have a plan. This plan or format will be a set of steps to be followed when working any problems involving a chemical equation. Applying this format faithfully will allow you to reduce even difficult sounding problems to their basic parts for solution. Following is the format you should allow after carefully reading the problem. 1. Write the balanced chemical reaction. Write out what the problem has asked for. (To find) 3. Write out what information is given in the problem (Givens) 4. Write out the steps you are going to take to solve the problem (Route) 5. Assemble the necessary conversion factors (Conversions) 6. Set up the problem and perform the calculations (Calculations) 7. Record your answer including the correct units. (Answer) Six of the seven steps have to do with setting the problem up, with one step or about 15% involving any math which will always be multiplication or division. Following are three example problems employing this format.
Example Problem (1) Hydrogen and Oxygen react to form water. How many grams of water can be produced by the reaction of 8.5 x 10 5 grams of Oxygen assuming that an excess amount of Hydrogen is present. Solution Step 1 The Equation H +0 H 0 Step To find number of grams of H 0 Step 3 Given 8.5x10 5 grams of 0 and sufficient amount of Hydrogen Step 4 Route grams 0 moles0 molesh 0 grams H 0 Step 5 Conversions 3g O = 1 mole O (from MW) 1 mole O = mole H O (from equation) g H O = 1 mole H O (from MW) Step 6 - Calculations Write the quantity given and transform it by means of the conversion factors set up as fractions. Let the units determine which is the numerator and denominator of the fractions. 8.5 105 g O 1mole O 3g O mole HO gho = 1mole O 1mole H O Canceling the units and performing the math gives Step 7 Answer 9.6 x 10 5 g H O ( 8.5 105 )( 1)( ) ( ) ( 3)( 1)( 1) g HO = 9.6 10 5 g H O
Example Problem () Solution Water reacts with poisonous carbon monoxide to form Hydrogen gas and Carbon Dioxide. This is called the water-gas reaction. How many pounds of water will be needed to react with 150 pounds of Carbon Monoxide? Step 1 Equation H O + CO H + CO Step To find number of pounds of water Step 3 Given 150 pounds of CO Step 4 Route lbs CO g CO moles CO lbs H O g H O moles H O Step 5 Conversions 1 lb = 453.6 g 8 g CO = 1 mole CO 1 mole CO = 1 mole H O g = 1 mole H O Step 6 Calculations 150 lbs CO 453.6g CO 1lb CO g HO 1lb HO = 804lbs HO 1mole H O 453.6g H O Step 7 Answer 804 lbs of water 1mole CO 1mole HO 8g CO 1mole CO
Example Problem (3) Solution In your automobile, gasoline (or octane) burns in Oxygen to produce Carbon Dioxide, water, and heat. This is why you sometimes see water coming out of an exhaust pipe or frozen water vapor trailing out of the exhaust on a cold winter day. How much water in gallons is produced by the burning of one gallon of gasoline? Step 1 Equation C 8 H + 5O 16CO + H O Step To find number of gallons of H O Step 3 Given one gallon of C 8 H Step 4 Route gal C 8 H mls C 8 H g C 8 H mole C 8 H gal H O mls H O g H O mole H O Step 5 Conversions 1 gal = 4 qt. 1 l = 1.057 qt. 1000 ml = 1 l 0.95 g = 1 ml (density of C 8 H ) 114 g C 8 H = 1 mole C 8 H mole C 8 H = mole H O g H O = 1 mole H O 1 g = 1 ml (density of H O) Step 6 Calculations 1gal C H 8 4 qt. 1gal 1l 10.57 qt. 1000 ml 0.95g CH 1l 1ml 1mole C8H 114g C H 8 mole C8H mole H O g HO 1ml HO 1l HO 1mole H O 1g H O 1000 ml 1.057 qt. 1l 1gal 4 qt. = 0.167 gal H O Step 7 Answer 1.67 x 10 - gallons of H O This problem was used to illustrate the format even though large numbers of conversions are necessary.
PROBLEMS 1. How many grams of NH 3 can be produced by 10 grams of N and 10 grams of H in the following reaction. 3H +N NH 3. How many grams each of Sulfur and Oxygen are needed to produce 15 tons of SO. S + 0 SO 3. How many pounds of oxygen are needed to react with 0 gallons of gasoline. C 8 H + 5 0 16C0 + H 0 4. How many moles of Nitrogen Dioxide can be produced by decomposition of 1.85 x 10 8 grams of Dinitrogen Tetraoxide N 0 4 N0 5. How much Nitric Oxide can be produced by reaction of one ton of air (80% N, 0% 0 by weight) N + 0 N0