# It’s time to learn about... Stoichiometry: Limiting Reagents At the conclusion of our time together, you should be able to: 1. Determine the limiting.

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It’s time to learn about...

Stoichiometry: Limiting Reagents At the conclusion of our time together, you should be able to: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

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Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed and 4 extra eggs, could we make 4 dozen cookies? If not, what limits us?? What if we only had one egg, could we make 3 dozen cookies???

Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. excess That reactant is said to be in excess (there is too much). limiting reactant The other reactant limits how much product we get. Once it runs out, the reaction ’s. This is called the limiting reactant.

Limiting Reactant: Example 10.0 g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.33 g AlCl 3 26.98 g Al 2 mol Al 1 mol AlCl 3 = 49.4 g AlCl 3 35.0 g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.33 g AlCl 3 70.90 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9 g AlCl 3 Limiting Reactant

LR Example Continued We get 49.4 g of aluminum chloride from the given amount of aluminum, but only 43.9 g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0 g of chlorine is used up, the reaction comes to a complete.

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Limiting Reactant Practice #5 3.00 g of magnesium reacts with 2.20 g of oxygen. Calculate which reactant is in excess and how much product is made. A little quicker way to do this is to pick one reactant and determine what amount of the other reactant is needed to completely react the first.

Calculate the mass in grams of oxygen required to react completely with 3.00 g of magnesium. Mg+ O 2  MgO 2 Mg+ O 2  2 MgO = 1.97 g O 2 x 1mol O 2 2 mol Mg x 32.00 g O 2 1 mol O 2 3.00 g Mg x 1 mol Mg 24.31 g Mg

Finding the Amount of Excess To completely react 3.00 g of magnesium, we would need 1.97 g of oxygen. Since we have 2.20 g of oxygen, the magnesium limits the amount of product we can make. Oxygen is in excess. By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Finding the Amount of Excess Can we find the amount of excess oxygen in the previous problem? To completely react 3.00 g of magnesium, we would need 1.97 g of oxygen. Since we have 2.20 g of oxygen, the magnesium limits the amount of product we can make. Oxygen is in excess. 2.20 g O 2 available – 1.97 g O 2 used = 0.23 g O 2 excess

3.00 g Mg Calculate the mass in grams of magnesium oxide produced when 3.00 g of magnesium (the limiting reagent) is burned in excess oxygen. Mg+ O 2  MgO 2 Mg+ O 2  2 MgO = 4.97 g MgO x 2 mol MgO 2 mol Mg x 40.31 g MgO 1 mol MgO x 1 mol Mg 24.31 g Mg

Gee, I wonder what’s going to happen next??

Stoichiometry: Limiting Reagents At the conclusion of our time together, you should be able to: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Pick a reactant (A), convert grams to moles, and compare it to the grams needed of the other reactant (B). 3. If the answer for B is higher than what you have in the problem, 4. You don’t have enough B, therefore B is limiting. 5. If the answer for B is lower than what you have in the problem, 6. You have enough B, therefore A is limiting. 7. If you have enough B, subtract your answer for B from what you have in the problem = excess B.

Stoichiometry: Limiting Reagents Let’s see if you can: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

Interesting Picture:

x 3 mol Zn 2 mol MoO 3 What mass of ZnO is formed when 20.0 g of MoO 3 is reacted with 10.0 g of Zn? 20.0 g MoO 3 x 1 mol MoO 3 143.94 g MoO 3 3 Zn + 2 MoO 3  Mo 2 O 3 + 3 ZnO x 65.39 g Zn 1 mol Zn = 13.6 g Zn

x 3 mol ZnO 3 mol Zn What mass of ZnO is formed when 20.0 g of MoO 3 is reacted with 10.0 g of Zn? We need 13.6 g of Zn so Zn is limiting! 10.0 g Znx 1 mol Zn 65.39 g Zn 3 Zn + 2 MoO 3  Mo 2 O 3 + 3 ZnO x 81.39 g ZnO 1 mol ZnO = 12.4 g ZnO

Stoichiometry: Limiting Reagents Let’s see if you can: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

x 1 mol NH 4 NO 3 80.06 g NH 4 NO 3 30.0 g NH 4 NO 3 Another Limiting Reagent Worksheet #1 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 a. Which is limiting? = 31.9 g NaNO 3 x 3 mol NaNO 3 3 mol NH 4 NO 3 x 85.00 g NaNO 3 1 mol NaNO 3

x 1 mol Na 3 PO 4 163.94 g Na 3 PO 4 50.0 g Na 3 PO 4 Another Limiting Reagent Worksheet #1 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 a. Which is limiting? = 77.8 g NaNO 3 x 3 mol NaNO 3 1 mol Na 3 PO 4 x 85.00 g NaNO 3 1 mol NaNO 3

x 1 mol NH 4 NO 3 80.06 g NH 4 NO 3 30.0 g NH 4 NO 3 a. Which is limiting? NH 4 NO 3 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 b. Maximum amount of each product? = 18.6 g (NH 4 ) 3 PO 4 x 1 mol (NH 4 ) 3 PO 4 3 mol NH 4 NO 3 x 149.12 g (NH 4 ) 3 PO 4 1 mol (NH 4 ) 3 PO 4

x 1 mol NH 4 NO 3 80.06 g NH 4 NO 3 30.0 g NH 4 NO 3 a. Which is limiting? NH 4 NO 3 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 b. Maximum amount of each product? = 31.9 g NaNO 3 x 85.00 g NaNO 3 1 mol NaNO 3 x 3 mol NaNO 3 3 mol NH 4 NO 3

x 1 mol NH 4 NO 3 80.06 g NH 4 NO 3 30.0 g NH 4 NO 3 Another Limiting Reagent Worksheet #1 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 a. Which is limiting? = 20.5 g Na 3 PO 4 x 1 mol Na 3 PO 4 3 mol NH 4 NO 3 x 163.94 g Na 3 PO 4 1 mol Na 3 PO 4

x 1 mol NH 4 NO 3 80.06 g NH 4 NO 3 30.0 g NH 4 NO 3 a. Which is limiting? NH 4 NO 3 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 b. Maximum amount of each product? = 18.6 g (NH 4 ) 3 PO 4 x 1 mol (NH 4 ) 3 PO 4 3 mol NH 4 NO 3 x 149.12 g (NH 4 ) 3 PO 4 1 mol (NH 4 ) 3 PO 4

x 1 mol NH 4 NO 3 80.06 g NH 4 NO 3 30.0 g NH 4 NO 3 a. Which is limiting? NH 4 NO 3 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 b. Maximum amount of each product? = 31.9 g NaNO 3 x 85.00 g NaNO 3 1 mol NaNO 3 x 3 mol NaNO 3 3 mol NH 4 NO 3

- 20.5 g NH 3 PO 4 50.0 g Na 3 PO 4 c. How much of the other reagent is left over?? 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 = 29.5 g NH 3 PO 4

Stoichiometry: Limiting Reagents Let’s see if you can: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

FeS Lab Problem: In the lab where you combined iron (II) with sulfur, if 5.00 g of iron is combined with 5.00 g of sulfur, what is the limiting reagent and how much excess reagent is left? 5.00 g Fex 1 mol Fe 55.85 g Fe Fe + S  FeS x 1 mol S I mol Fe = 2.87 g Sx 32.07 g S 1 mol S

When 5.00 g of Fe is used, I would need 2.87 grams of S to completely react the Fe. Since I have 5.00 g of S, I have plenty of S and therefore, the Fe is limiting. 2.87 g S used 5.00 g S – 2.87 g S used = 2.13 g excess S

Stoichiometry: Limiting Reagents with Percent Yield At the conclusion of our time together, you should be able to: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction 4. Determine the percent yield for a problem

What is the Percent Yield of a Chemical Reaction? Actual Product Produced Theoretical Product that Should be Produced X 100 = Percent Yield

Limiting Reactant Practice 15.0 g of aluminum reacts with 15.0 g of iodine. Calculate which reactant is limiting, how much excess reactant is available and how much product is made. Determine the percent yield if 10.5 g of aluminum iodide is actually produced.

Calculate the mass in grams of iodine required to react completely with 15.0 g of aluminum. Al + I 2  AlI 3 2 Al + 3 I 2  2 AlI 3 = 212 g I 2 x 3 mol I 2 2 mol Al x 253.80 g I 2 1 mol I 2 15.0 g Al x 1 mol Al 26.98 g Al

Finding the Amount of Excess To completely react 15.0 g of aluminum, we would need 212 g of iodine. Since we only have 15.0 g of iodine, the iodine limits the amount of product we can make. By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Finding Excess Practice 15.0 g of aluminum reacts with 15.0 g of iodine. Calculate the excess of aluminum. 2 Al + 3 I 2  2 AlI 3 Always start with the limiting reactant: 15.0 g I 2 1 mol I 2 2 mol Al 26.98 g Al 253.80 g I 2 3 mol I 2 1 mol Al = 1.06 g Al USED! 15.0 g Al – 1.06 g Al = 13.9 g Al EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only use it!

x 1 mol I 2 253.80 g I 2 15.0 g I 2 Calculate the mass in grams of aluminum iodide that would be produced in this reaction. Al + I 2  AlI 3 2 Al + 3 I 2  2 AlI 3 = 16.1 g AlI 3 x 2 mol AlI 3 3 mol I 2 x 407.68 g AlI 3 1 mol I 2

10.5 g AlI 3 16.1 g AlI 3 Determine the percent yield if 10.5 g of aluminum iodide is produced. 16.1 g of AlI 3 should have been produced. What is the percent yield? 2 Al + 3 I 2  2 AlI 3 = 65.2 % Yield x 100

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Stoichiometry: Limiting Reagents with Percent Yield Let’s see if you can: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction 4. Determine the percent yield for a problem

Percent Yield #1 How many grams of antimony(III) iodide would be produced? Determine the percent yield if 118.00 g of antimony(III) iodide is actually produced.

253.80 g I 2 x 1 mol I 2 98.60 g I 2 How many grams of antimony(III) iodide would be produced? Sb + I 2  SbI 3 2 Sb + 3 I 2  2 SbI 3 = 130.1 g SbI 3 x 2 mol SbI 3 3 mol I 2 x 502.46 g SbI 3 1 mol SbI 3

x 118.00 g SbI 3 130.1 g SbI 3 Determine the percent yield if 118.00 g of antimony (III) iodide is produced. 130.1 g of Sbl 3 should have been produced. What is the percent yield? 2 Sb + 3 I 2  2 SbI 3 = 90.70 % Yield x 100

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