# Ch. 9 Notes -- Stoichiometry

## Presentation on theme: "Ch. 9 Notes -- Stoichiometry"— Presentation transcript:

Ch. 9 Notes -- Stoichiometry
Stoichiometry refers to the calculations of chemical quantities from __________________ chemical equations. Interpreting Everyday Equations 2 guards 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 2G 2F C G2F2C 14 16 7 guards

Interpreting Chemical Equations
___N2 (g) + ___H2 (g)  ___NH3 (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 ____ moles H2  ____ moles NH3 ____ molecule N ____ molecules H2  ____ molecules NH3 ____ liter N ____ liters H2  ____ liters NH3 ____ grams N ____ grams H2  ____ grams NH3 1 3 2 balance 1 3 2 1 3 2 1 3 2 28.0 6.0 34.0

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: N2 (g) H2 (g)  2NH3 (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 2 mol NH3 7 mol N2 14 moles of NH3 x = 1 mol N2 3 mol H2 8 mol N2 24 moles of H2 x = 1 mol N2 3 mol H2 15 moles of H2 10 mol NH3 x = 2 mol NH3

The Ultimate Stoichiometric Problem of Ultimate Destiny
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. N2 (g) H2 (g)  2NH3 (g) Practice Problem: How many grams of ammonia can be made from reacting 39.0 grams of nitrogen with an excess of hydrogen? 1 mol N2 2 mol NH3 17.0 g NH3 39.0 g N2 47.4 g NH3 x x x = 28.0 g N2 1 mol N2 1 mol NH3 The Ultimate Stoichiometric Problem of Ultimate Destiny

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. N2 (g) H2 (g)  2NH3 (g) Practice Problem: How many grams of ammonia can be made from reacting 20.0 liters of hydrogen with an excess of nitrogen? 1 mol H2 2 mol NH3 17.0 g NH3 20.0 L H2 10.1 g NH3 x x x = 22.4 L H2 3 mol H2 1 mol NH3

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. N2 (g) H2 (g)  2NH3 (g) Practice Problem: How many liters of ammonia can be made from reacting 36.0 grams of nitrogen with an excess of hydrogen? 1 mol N2 2 mol NH3 22.4 L NH3 36.0 g N2 57.6 L NH3 x x x = 28.0 g N2 1 mol N2 1 mol NH3

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. N2 (g) H2 (g)  2NH3 (g) Practice Problem: How many liters of ammonia can be made from reacting 125 liters of hydrogen with an excess of nitrogen? mole-mole 2 L NH3 125 L H2 83.3 L NH3 x = 3 L H2 …the long way to get the same answer… 1 mol H2 2 mol NH3 22.4 L NH3 125 L H2 83.3 L NH3 x x x = 22.4 L H2 3 mol H2 1 mol NH3

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

Practice Problems: N2 (g) + 3H2 (g)  2NH3 (g)
Calculations Step 1: Do a mole-mole conversion for one of the substances. This answer is how much of it you need. Step 2: Compare your answer to how much reactant was given. Do you have enough? If not, this reactant is your limiting reagent! Practice Problems: N2 (g) H2 (g)  2NH3 (g) 1) If 2.7 moles of nitrogen reacts with 6.3 moles of hydrogen, which will you run out of first? 2) If 3.9 moles of nitrogen reacts with 12.1 moles of hydrogen, what is the limiting reagent? Were you given enough H2? ______ Limiting Reagent = ______ 3 mol H2 No ! (6.3<8.1) 2.7 mol N2 8.1 moles of H2 are needed for the reaction. x = 1 mol N2 H2 Were you given enough H2? ______ Limiting Reagent = ______ 3 mol H2 Yes ! (12.1>11.7) 3.9 mol N2 11.7 moles of H2 are needed for the reaction. x = 1 mol N2 N2

Excess Reagent (or Excess Reactant)
How many moles of excess reagent do you have? Step 1: Do a mole-mole conversion starting with the limiting reagent as the given. The answer you get is how much of the excess reagent you need to completely react with the limiting reagent. Sometimes you get lucky and you already did this conversion from the previous problem! Step 2: Subtract this answer from the amount given in the original problem, and that is how many moles of excess reagent there are.

Excess Reagent (or Excess Reactant)
Practice Problem: Find the number of moles of excess reagent from the previous practice problems. 1 mol N2 6.3 mol H2 2.1 moles of N2 are needed for the reaction. x = 3 mol H2 2.7 moles of N2 were originally given, so the excess will be… 2.7 moles given − 2.1 moles needed = 0.6 moles of N2 excess For the second practice problem, we already started with the limiting reagent, so all you have to do is subtract… 12.1 moles given −11.7 moles needed= 0.4 moles of H2 excess

Percent Yield 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! efficient x 100 % Yield = actual theoretical

Percent Yield Practice Problem: 2H2 (g) + O2 (g)  2H2O (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! 1 mol H2 2 mol H2O 18.0 g H2O 40 g H2 360 g H2O x x x = 2.0 g H2 2 mol H2 1 mol H2O 300 % Yield = x 100 = 83.3% 360

Theoretical Yield = 2.0 g + 16.0 g = 18.0 grams
Percent Yield Practice Problem: H2 (g) O2 (g)  2H2O (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!) Theoretical Yield = 2.0 g g = 18.0 grams 17.5 % Yield = x 100 = 97.2% 18.0