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Ch. 9 Notes – Chemical Quantities Stoichiometry refers to the calculations of chemical quantities from __________________ chemical equations. Interpreting.

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Presentation on theme: "Ch. 9 Notes – Chemical Quantities Stoichiometry refers to the calculations of chemical quantities from __________________ chemical equations. Interpreting."— Presentation transcript:

1 Ch. 9 Notes – Chemical Quantities Stoichiometry refers to the calculations of chemical quantities from __________________ chemical equations. Interpreting Everyday Equations 2 guards + 2 forwards + 1 center  1 basketball team ___ + ___ + ___  __________ Practice Problems: 1) How many guards does it take to make 7 teams? ______ 2) How many forwards are there in 8 teams? ______ 3) If you have 8 centers, 17 forwards and 14 guards, how many teams can be made? ______ What do you run out of first? _________ balanced 2G2FCG2F2CG2F2C guards

2 Interpreting Chemical Equations ___N 2 (g) + ___H 2 (g)  ___NH 3 (g) The first thing that must be done is to ______________ the equation! Here are the kinds of information you can get from the equation: ____ mole N 2 + ____ moles H 2  ____ moles NH 3 ____ molecule N 2 + ____ molecules H 2  ____ molecules NH 3 ____ liter N 2 + ____ liters H 2  ____ liters NH 3 ____ grams N 2 + ____ grams H 2  ____ grams NH 3 32 balance

3 Mole-Mole Conversions The mole conversion factor comes from the _________________ of the balanced chemical equation. Step 1: Write down the “given”. Step 2: Set up a conversion factor to change from moles to moles. Practice Problems: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1) How many moles of ammonia can be made from 7 moles of nitrogen reacting with an excess of hydrogen? 2) How many moles of hydrogen are required to completely react with 8 moles of nitrogen to produce ammonia? 3) How many moles of hydrogen are needed to react with an excess of nitrogen to make 10 moles of ammonia? coefficients 1 mol N 2 2 mol NH 3 7 mol N 2 x= 14 moles of NH 3 1 mol N 2 3 mol H 2 8 mol N 2 x= 24 moles of H 2 2 mol NH 3 3 mol H 2 10 mol NH 3 x= 15 moles of H 2

4 Other Conversion Problems Mass-Mass: (grams to moles to moles to grams) Step 1: Write down the “given” and convert from grams to moles. Step 2: Convert from moles of the “given” to moles of the “unknown” using a mole conversion factor. Step 3: Convert from moles of the unknown to grams. N 2 (g) + 3H 2 (g)  2NH 3 (g) Practice Problem: How many grams of ammonia can be made from reacting 39.0 grams of nitrogen with an excess of hydrogen? 28.0 g N 2 2 mol NH 3 1 mol N g N 2 x x = 47.4 g NH 3 1 mol N 2 x 1 mol NH g NH 3

5 Other Conversion Problems Volume-Mass: (liters to moles to moles to grams) Step 1: Write down the “given” and convert from liters to moles. Step 2: Convert from moles of the “given” to moles of the “unknown” using a mole conversion factor. Step 3: Convert from moles of the unknown to grams. N 2 (g) + 3H 2 (g)  2NH 3 (g) Practice Problem: How many grams of ammonia can be made from reacting 20.0 liters of hydrogen with an excess of nitrogen? 22.4 L H 2 2 mol NH 3 1 mol H L H 2 x x = 10.1 g NH 3 3 mol H 2 x 1 mol NH g NH 3

6 Other Conversion Problems Mass-Volume: (grams to moles to moles to liters) Step 1: Write down the “given” and convert from grams to moles. Step 2: Convert from moles of the “given” to moles of the “unknown” using a mole conversion factor. Step 3: Convert from moles of the unknown to liters. N 2 (g) + 3H 2 (g)  2NH 3 (g) Practice Problem: How many liters of ammonia can be made from reacting 36.0 grams of nitrogen with an excess of hydrogen? 28.0 g N 2 2 mol NH 3 1 mol N g N 2 x x = 57.6 L NH 3 1 mol N 2 x 1 mol NH L NH 3

7 Other Conversion Problems Volume-Volume: (treat it exactly like a ____________ conversion) Step 1: Write down the “given”. Step 2: Convert from liters to liters using a mole conversion factor. N 2 (g) + 3H 2 (g)  2NH 3 (g) Practice Problem: How many liters of ammonia can be made from reacting 125 liters of hydrogen with an excess of nitrogen? 3 L H 2 2 L NH L H 2 x= 83.3 L NH L H 2 2 mol NH 3 1 mol H L H 2 x x = 83.3 L NH 3 3 mol H 2 x 1 mol NH L NH 3 …the long way to get the same answer… mole-mole

8 The limiting reagent is the reactant that ___________ _____ first. The reactant that is in abundance is called the ___________ reagent. Limiting Reagent (or Limiting Reactant) runs out excess

9 Calculations Step 1: Do a conversion for both given reactants into a product. Step 2: The smaller amount of product was made by the limiting reactant. The smaller amount would be the actual amount produced. Practice Problems: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1) If 2.7 moles of nitrogen reacts with 6.3 moles of hydrogen, which is your limiting reactant? How much NH 3 is produced? 1 mol N 2 2 mol NH mol N 2 x= 5.4 moles of NH 3 3 mol H 2 2 mol NH mol H 2 x= 4.2 moles of NH 3 My limiting reactant is H 2 and 4.2 moles of NH 3 will be produced.

10 Calculations Practice Problems: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1) If 37 grams of nitrogen reacts with 55 grams of hydrogen, which is your limiting reactant? How much product is produced? 28.0 g N 2 2 mol NH 3 1 mol N g N 2 x x = 44.9 g NH 3 1 mol N 2 x 1 mol NH g NH g H 2 2 mol NH 3 1 mol H g H 2 x x = g NH 3 3 mol H 2 x 1 mol NH g NH 3 My limiting reactant is N 2 and 44.9g of NH 3 will be produced.

11 How many grams of excess reagent do you have? Step 1: Convert each reactant into product to find the limiting reactant. Step 2: Subtract the two answers to find the amount of product that was not made. Step 3: Convert this amount of product back into the excess reactant. This is the amount of excess reagent not used in the reaction. Excess Reagent (or Excess Reactant)

12 Practice Problem: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1) If 37 grams of nitrogen reacts with 55 grams of hydrogen, how much product is made and much excess reactant is left over? Excess Reagent (or Excess Reactant) 28.0 g N 2 2 mol NH 3 1 mol N g N 2 x x = 44.9 g NH 3 1 mol N 2 x 1 mol NH g NH g H 2 2 mol NH 3 1 mol H g H 2 x x = g NH 3 3 mol H 2 x 1 mol NH g NH g – g = g NH g NH 3 1 mol H 2 1 mol NH g NH 3 x x = 6.36 g H 2 2 mol NH 3 x 1 mol H g H 2

13 Percent Yield is a ratio that tells us how ________________ a chemical reaction is. The higher the % yield, the more efficient the reaction is. Actual or “experimental” Yield Theoretical or “ideal” Yield The ___________ yield is the amount you experimentally get when you run the reaction in a lab. The _______________ yield is the amount you are ideally supposed to get if everything goes perfectly. You can calculate this amount using stoichiometry! Percent Yield x 100 % Yield = efficient actual theoretical

14 Practice Problem: 2H 2 (g) + O 2 (g)  2H 2 O (g) 1) A student reacts 40 grams of hydrogen with an excess of oxygen and produces 300 grams of water. Find the % yield for this reaction. Step 1: Do a mass-mass conversion starting with the given reactant and converting to the product, (in this example, the water.) The answer you get is how much water you “theoretically” should have produced. Step 2: The other value in the question, (300 grams), is what you actually produced. Divide them to get your % yield! Percent Yield 2.0 g H 2 2 mol H 2 O1 mol H 2 40 g H 2 x x = 360 g H 2 O 2 mol H 2 x 1 mol H 2 O 18.0 g H 2 O % Yield = x 100 =83.3%

15 Practice Problem: 2H 2 (g) + O 2 (g)  2H 2 O (g) 2) If 2.0 grams of hydrogen completely reacted with 16.0 grams of oxygen but only produced 17.5 grams of water, what is the % yield for the reaction? (Since more information is given in this question, we won’t necessarily have to do the conversion for Step 1!) Percent Yield % Yield = x 100 =97.2% Theoretical Yield = 2.0 g g = 18.0 grams


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