Supplemental Activities. Module: States of Matter. Section: Reaction Stoichiometry Key

Transcription

1 Supplemental Activities Module: States of Matter Section: Reaction Stoichiometry Key

2 Balancing Chemical Reactions Activity 1 The purpose of this activity is to check your understanding of the concept Law of Conservation of mass and apply that to balancing chemical equations. 1. How does the Law of Conservation of Mass help us balance chemical equations? The Law of Conservation of Mass tells us that there should not be a significant loss of mass across a chemical change. In this way we can say with relative certainty that if there were five carbon atoms in the reactants there will be five carbon atoms in the products. They may be arranged differently, but they will not have disappeared or multiplied. This principle helps us count up atoms on either side of a chemical reaction and balance them properly. 2. Balance the following equation: 1C 6H 12O 6 + 6O 2 6CO 2 + 6H 2O 3. Balance the following equation: 1Co 2 + 3H 2 2Co + 3H 2S 4. Balance the following equation: 2H 3PO 4 + 3Ca(OH) 2 6H 2O + 1Ca 3(PO 4) 2 Activity 2 The purpose of this activity is to check if you remember how to name compounds! 1. Covalent compounds contain non-metals. Ionic compounds often contain a metal and a nonmetal although sometimes they are formed using one or more of the polyatomic ions that a good student of chemistry should be able to recognize. 2. Name each of the compounds found in the chemical equations in question 3 in the previous activity. Co 2 cobalt (III) sulfide H 2 dihydrogen (molecular hydrogen) Co cobalt H 2S dihydrogen monosulfide

3 3. Write down and balance this chemical reaction: magnesium nitride reacts with water to give magnesium hydroxide and nitrogen trihydride (ammonia). 1Mg 3N 2 + 6H 2O 3Mg(OH) 2 + 2NH 3 Mole to Mole Conversions ACTIVITY 1 The purpose of this activity is to check your understanding of the concept of mole-to-mole conversions across a chemical change. 1. How many moles of carbon dioxide are formed when 3.7 moles of C 3 reacts with excess oxygen to form CO 2 and H 2O according to the following balanced chemical reaction: C 3 + 5O 2 3CO 2 + 4H 2O The balanced chemical equation provides the key for converting between C 3 and CO 2. The of C 3 to CO 2 is 1 to 3. mol C 3 mol CO mol C 3 3 mol CO 2 1 mol C 3 =11.1 mol CO 2 11 mol CO 2 ACTIVITY 2 The purpose of this activity is to check your ability to convert between moles of one reactant given and moles of other reactant needed. 1. How many moles of Fe are necessary to react completely with 0.91 moles of O 2 given the following balanced chemical reaction: 4Fe + 3O 2 2Fe 2 The balanced chemical equation provides the key for converting between O 2 and Fe. The of O 2 to Fe is 3 to 4. We can use this to determine how many moles of iron were needed to react completely with 0.91 moles of O 2. mol O 2 mol Fe 0.91 mol O 2 4 mol Fe 3 mol O 2 =1.213 mol Fe 1.2 mol Fe

4 2. For the same balanced equation and assuming you have plenty of iron, how many moles of O 2 would be necessary to form 22 moles of Fe 2? The balanced chemical equation provides the key for converting between Fe 2 and O 2. The of Fe 2 to O 2 is 2 to 3. We can use this to determine how many moles of O 2were needed to produce 22 moles of Fe 2. mol Fe 2 mol O 2 22 mol Fe 2 3 mol O 2 2 mol Fe 2 = 33 mol O 2 Activity 3 The purpose of this activity is to check your ability to determine the limiting reactant and from there correctly predict the amount of product formed. 1. Given 2.1 moles of Al 2 and 27 moles of H 2O how many moles of H 2S would be formed according to the following balanced chemical equation: Al 2 + 6H 2O 2Al(OH) 3 + 3H 2S There are a few methods to determine the limiting reactant. We could recognize that the balanced equation indicates that the reaction requires six times the amount of H 2O as it does Al 2 and since 27 moles is more than six times 2.1 moles, then H 2O is likely not the limiting reactant. We could run the stoichiometric analysis twice and see which reactant produces the fewest moles of product. The reactant that produces the fewest moles of product is the limiting reactant. We could also divide the moles of reactant given by their respective stoichiometric coefficients. The smallest quotient is the limiting reactant.

5 Calculating moles H 2 S twice : 2. 3 mol H 2 S = 6.3 mol H 2 S 27 mol H 2 O 3 mol H 2 S O =13.5 mol H 2S 6.3 mol H 2 S <13.5 mol H 2 S Al 2 is L.R. Dividing by stoichiometric coefficients: 2. = mol H 2 O O = < 4.5 Al 2 is L.R. so mol H 2 S = 6.3 mol H 2 S Mass to Mass conversions ACTIVITY 1 The purpose of this activity is to check your ability to convert between mass of one reactant given and mass of product formed. 1. How many grams of water are formed when 63.0 g of C 3 reacts with excess oxygen to form CO 2 and H 2O according to the following balanced chemical reaction: C 3 + 5O 2 3CO 2 + 4H 2O molar mass H 2 O molar mass C grams C 3 H 3 8 mol C 3 mol H 2 O grams H 2 O 63.0 g C 3 1 mol C g C 3 4 mol H 2 O 1 mol C g H 2 O 1 mol H 2 O =103 g H 2 O

6 2. Assuming you have plenty of iron, how many grams of O 2 would be necessary to form 3.7 g of Fe 2 according to the following balanced chemical equation: 4Fe + 3O 2 2Fe 2 molar mass Fe grams Fe 2 O 2 molar mass O 3 mol Fe 2 mol O 2 2 grams O g Fe 2 1 mol Fe g Fe 2 3 mol O 2 2 mol Fe g O 2 1 mol O 2 =1.1 g O 2 ACTIVITY 2 The purpose of this activity is to check your ability to convert between mass of reactant given and mass of other reactant needed to form product. 1. How many grams of Al 2 are necessary to react completely with 29.2 g of H 2O given the following balanced chemical reaction: Al 2 + 6H 2O 2Al(OH) 3 + 3H 2S molar mass H 2 O molar mass Al grams H 2 O mol H 2 O mol Al 2 S 2 3 grams Al g H 2 O 1 mol H 2O 18.0 g H 2 O O 150 g Al 2 = 40.6 g Al 2 ACTIVITY 3 The purpose of this activity is to check your ability to determine the limiting reactant and from there correctly predict the amount of product formed. 1. Given 28.7 g of Al 2 and 83.9 g of H 2O how many grams of Al(OH) 3 would be formed according to the following balanced chemical reaction: Al 2 + 6H 2O 2Al(OH) 3 + 3H 2S To compare between different compounds, we must compare them in moles. So first we need to convert the mass amounts of both compounds into moles. Then, we can use any method to determine which reactant is the limiting reactant and how many grams of Al(OH) 3 would be formed.

7 Calculating moles Al(OH) 3 twice : 28.7 g Al g Al 2 2 mol Al(OH) g Al(OH) 3 1 mol Al(OH) 3 = 29.8 g Al(OH) g H 2 O 1 mol H 2 O 18.0 g H 2 O 2 mol Al(OH) 3 O 78.0 g Al(OH) 3 1 mol Al(OH) 3 =121 g Al(OH) g Al(OH) 3 <121 g Al(OH) 3 Al 2 is L.R. Dividing by stoichiometric coefficients: 28.7 g Al g Al 2 = = g H 2 O 1 mol H 2 O 18.0 g H 2 O = 4.6 O 4.6 O O = < Al 2 is L.R. so g Al g Al 2 2 mol Al(OH) g Al(OH) 3 1 mol Al(OH) 3 = 29.8 g Al(OH) 3 Percent Yield Activity 1 The purpose of this activity is for you check your understanding of percent yield. 1. The actual yield of a particular reaction was grams whereas the theoretical yield was computed to be g. What is the percent yield?

8 Actual %Yield = Theoretical 100% %Yield = 67.50g 89.88g 100% %Yield = 75.10% 2. Imagine that you walk into the first day of chemistry lab and your TA informs you that you will be calculating the percent yield of a compound produced in a reaction. She explains that you should be able to provide a theoretical yield within five minutes and by the end of the lab session you should be able to provide the actual yield as well as the percent yield. As an excellent chemistry student you readily agree and start preparing the theoretical yield, however your lab partner thinks you should take a shortcut and just provide all three numbers now. Explain to your lab partner how you would provide the values for theoretical, actual and percent yield within your TA s time limits. The theoretical yield can be calculated quickly because we use the balanced chemical equation to calculate how much product we could theoretically form. To obtain the actual yield though we have to actually perform the reaction and weigh the final product, which will take a good amount of time during the lab session. Finally, to calculate the percent yield we need both the theoretical and actual yields. We divide the actual by the theoretical and multiply by 100 to find the percent yield. Therefore, we can t take a short cut and calculate all three numbers right now! Activity 2 1. The percent yield of CO 2 for the reaction C 3 + 5O 2 3CO 2 + 4H 2O is 68%. What mass of CO 2 is expected from the reaction of 6.5 g of propane (C 3) with an excess of oxygen? What mass of CO 2 did the reaction produce in this experiment? Most importantly we need to check that we are given a balanced chemical equation (which we are). Next, we should calculate the amount of theoretical yield: 6.5 g C 3 1 mol C 3 44 g C 3 3 mol CO 2 1 mol C 3 44 g CO 2 1 mol CO 2 =19.5 g CO 2 Now, we can use the percent yield equation to determine the mass of PCl 5 that was actually collecting during the experiment:

9 Actual %Yield = Theoretical 100% Actual 68% = 100% 19.5 g CO 2 Actual 0.68 = 19.5 g CO 2 Actual =13.26 g CO 2 13g CO 2 2. Your lab partner from the previous activity (problem #2) is upset that your reaction didn t have a 100% percent yield. Explain to your lab partner why a 100% yield would be surprising for most reactions. There are many reasons why a reaction may not produce the full theoretical yield. Temperature and pressure can often affect the yield of a reaction. Impurities and sidereactions can use up reactants or obscure the product. Aim for a high percent yield of product, but don t be discouraged if you don t reach 100% yield.