Jeopardy $100 Stoich Vocab Stoich Calculations Stoich True/False Percent Yield Limiting Reactants $200 $300 $400 $500 $400 $300 $200 $100 $500 $400 $300 $200 $100 $500 $400 $300 $200 $100 $500 $400 $300 $200 $100 Final Jeopardy Final Jeopardy

Vocab - $100 This word describes the study of relationships between the amount of reactants used and products formed in a chemical reaction. This word describes the study of relationships between the amount of reactants used and products formed in a chemical reaction. Stoichiometry Stoichiometry

Vocab - $200 The ratio between the numbers of moles of any two substances in a balanced chemical equation. The ratio between the numbers of moles of any two substances in a balanced chemical equation. Mole Ratio Mole Ratio

Vocab - $300 This term describes the number of grams of a substance in 1 mol of the same substance. This term describes the number of grams of a substance in 1 mol of the same substance. Molar mass Molar mass

Vocab - $400 This term describes the reactant that is used up first in a chemical reaction. This term describes the reactant that is used up first in a chemical reaction. Limiting Reactant Limiting Reactant

Vocab - $500 This term describes the amount of a substance that should be produced in a chemical reaction. Found by doing a stoichiometric calculation. This term describes the amount of a substance that should be produced in a chemical reaction. Found by doing a stoichiometric calculation. Theoretical Yield Theoretical Yield

Stoich Calculations - $100 This conversion factor should be used when doing Mole to Mole Stoich Calculations This conversion factor should be used when doing Mole to Mole Stoich Calculations Mole Ratio Mole Ratio

Stoich Calculations - $200 2 NO + 1 O 2 2NO 2 2 NO + 1 O 2 2NO 2 How many moles O 2 will be needed to produce 3.6 moles of NO 2 ? How many moles O 2 will be needed to produce 3.6 moles of NO 2 ? 1.8 mol O mol O 2

Stoich Calculations - $300 This conversion factor should be used when calculating moles of a substance when given grams of the same substance. This conversion factor should be used when calculating moles of a substance when given grams of the same substance. Molar Mass (g/1 mol) Molar Mass (g/1 mol)

Stoich Calculations - $400 1 C 3 H O 2 3 CO H 2 O 1 C 3 H O 2 3 CO H 2 O How many moles of C 3 H 8 are required to produce 83.5 grams of CO 2 ? How many moles of C 3 H 8 are required to produce 83.5 grams of CO 2 ? mol C 3 H mol C 3 H 8

Stoich Calculations - $500 1 C 3 H O 2 3 CO H 2 O 1 C 3 H O 2 3 CO H 2 O How many grams of C 3 H 8 are required to produce 7.60 moles of CO 2 ? How many grams of C 3 H 8 are required to produce 7.60 moles of CO 2 ? 112 g C 3 H g C 3 H 8

True/False - $100 Molar mass is the only conversion factor needed in Mole to Mole stoich calculations. Molar mass is the only conversion factor needed in Mole to Mole stoich calculations. False, only need to use Mole Ratio False, only need to use Mole Ratio

True/False - $200 Stoichiometry is based on the law of conservation of mass, which is why all stoich calculations must start with a balanced equation. Stoichiometry is based on the law of conservation of mass, which is why all stoich calculations must start with a balanced equation. True True

True/False - $300 Experimental yield is the amount in grams found by doing a stoich calculation. Experimental yield is the amount in grams found by doing a stoich calculation. False, experimental yield is the actual amount produced by doing a laboratory experiment. Theoretical yield is the amount that should be produced “on paper.” False, experimental yield is the actual amount produced by doing a laboratory experiment. Theoretical yield is the amount that should be produced “on paper.”

True/False - $400 In a mass to mass stoich calculation, the following steps are required. In a mass to mass stoich calculation, the following steps are required. 1- molar mass of given substance 1- molar mass of given substance 2- mole ratio from balanced equation 2- mole ratio from balanced equation 3- molar mass of wanted substance 3- molar mass of wanted substance True True

True/False - $500 If you were making peanut butter and jelly sandwiches and ran out of jelly first, then jelly would be considered the excess reactant. If you were making peanut butter and jelly sandwiches and ran out of jelly first, then jelly would be considered the excess reactant. False, jelly would be the limiting reactant because it is used up first. Peanut butter and bread would be excess reactants in this case. False, jelly would be the limiting reactant because it is used up first. Peanut butter and bread would be excess reactants in this case.

Percent Yield - $100 What is the equation used to find percent yield? What is the equation used to find percent yield? Experimental yield/theoretical yield x 100 Experimental yield/theoretical yield x 100

Percent Yield - $200 Calculate the percent yield if 86.5 grams should have been produced, but only 74.6 were actually produced in a laboratory experiment. Calculate the percent yield if 86.5 grams should have been produced, but only 74.6 were actually produced in a laboratory experiment. 74.6/86.5 x 100 = 86% 74.6/86.5 x 100 = 86%

Percent Yield - $300 Describe how you could determine experimental yield in a laboratory setting. (Think about the stoichiometry lab and how you determined amount of copper produced) Describe how you could determine experimental yield in a laboratory setting. (Think about the stoichiometry lab and how you determined amount of copper produced) 1- Record Mass of empty beaker 1- Record Mass of empty beaker 2- Record Mass of substance and beaker after reaction has taken place 2- Record Mass of substance and beaker after reaction has taken place 3- Take Difference between initial and final mass to determine amount produced. 3- Take Difference between initial and final mass to determine amount produced.

Percent Yield - $400 If the percent yield is 76.4% and the theoretical yield is 45.7 grams, what is the experimental yield? If the percent yield is 76.4% and the theoretical yield is 45.7 grams, what is the experimental yield? Experimental yield = (.764 x 45.7) Experimental yield = (.764 x 45.7) Experimental yield = 34.9 grams Experimental yield = 34.9 grams

Percent Yield - $500 2 AgNO 3 + K 2 CrO 4 Ag 2 CrO KNO 3 2 AgNO 3 + K 2 CrO 4 Ag 2 CrO KNO 3 Determine the theoretical yield of Ag 2 CrO 4 when starting with g of AgNO 3. Determine the theoretical yield of Ag 2 CrO 4 when starting with g of AgNO 3. If the experimental yield is g, calculate the percent yield of the reaction. If the experimental yield is g, calculate the percent yield of the reaction. Theoretical yield = g Ag 2 CrO 4 Theoretical yield = g Ag 2 CrO 4 Percent yield = 0.455/0.488 x 100 = 93.2 % Percent yield = 0.455/0.488 x 100 = 93.2 %

Limiting Reactants - $100 N H 2 2 NH 3 N H 2 2 NH 3 Draw the reactants and product of this reaction in order to show a perfect stoichiometric mix. Draw the reactants and product of this reaction in order to show a perfect stoichiometric mix.

Limiting Reactants - $200 N H 2 2 NH 3 N H 2 2 NH 3 Draw the reactants and product of this reaction in order to show hydrogen gas as the limiting reactant. Draw the reactants and product of this reaction in order to show hydrogen gas as the limiting reactant.

Limiting Reactants - $300 Explain why the limiting reactant determines how much product can be produced in a chemical reaction. Explain why the limiting reactant determines how much product can be produced in a chemical reaction. Limiting reactant is the one that is used up first. If we run out of an ingredient needed to make the product, then no more product can be made. Limiting reactant is the one that is used up first. If we run out of an ingredient needed to make the product, then no more product can be made.

Limiting Reactants - $400 N H 2 2 NH 3 N H 2 2 NH mole of N 2 and 8.0 mole of H 2 are combined in a chemical reaction. Which is the limiting reactant and how much product will be formed? 4.0 mole of N 2 and 8.0 mole of H 2 are combined in a chemical reaction. Which is the limiting reactant and how much product will be formed? H 2 is the limiting reactant. 5.3 mole of NH 3 will be produced. H 2 is the limiting reactant. 5.3 mole of NH 3 will be produced.

Limiting Reactants - $500 Create a situation in order to show the difference between limiting and excess reactants. Create a situation in order to show the difference between limiting and excess reactants. Various Answers accepted Various Answers accepted

Final Jeopardy C 6 H 12 + O 2 CO 2 + H 2 O C 6 H 12 + O 2 CO 2 + H 2 O What type of reaction is shown above? What type of reaction is shown above? Balance the equation. Balance the equation. How many molecules of O 2 will be needed in order to produce 58.4 grams of H 2 O? How many molecules of O 2 will be needed in order to produce 58.4 grams of H 2 O? Combustion Reaction Combustion Reaction C 6 H O 2 6 CO H 2 O C 6 H O 2 6 CO H 2 O 2.93 x molecules of O x molecules of O 2