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1 Lab “Oreo Lab” 1. 6.3 Limiting Reagents Caution: this stuff can be difficult to follow at first. Be patient.

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Presentation on theme: "1 Lab “Oreo Lab” 1. 6.3 Limiting Reagents Caution: this stuff can be difficult to follow at first. Be patient."— Presentation transcript:

1 1 Lab “Oreo Lab” 1

2 6.3 Limiting Reagents Caution: this stuff can be difficult to follow at first. Be patient.

3 3 Limiting Reagents and Reagents in Excess So far we have assumed that all of the reactants taking part in the reaction have been transformed into products. (No left-overs) Not usually the case. 3

4 4 \ Ex. Wood burning –Unlimited oxygen in the air available for the reaction (oxygen = reagent in excess) –When all wood is burned, the reaction stops, no longer using oxygen from the air, and making no more water or CO 2. - The amount of wood limits the amount of products, therefor, it is the limiting reagent

5 5 Ex. Cake baking - The recipe calls for 2 eggs to combine with 1 cup of flour and this make 1 cake. - How many cakes could you bake if you have 4 eggs in your fridge, and 4 cups of flour in your pantry? - To use up all 4 eggs, you would only need 2 cups of flour. (2 cups left over). - Flour = reagent in excess - Eggs = limiting reagent 5

6 In Chemistry: Q - How many moles of NO are produced if __ mol NH 3 are burned in __ mol O 2 ? a) 4 mol NH 3, 5 mol O 2 b) 4 mol NH 3, 20 mol O 2 c) 8 mol NH 3, 20 mol O 2 In all of these cases, NH 3 limits the production of NO; if there was more NH 3, more NO would be produced Thus, NH 3 is called the “limiting reagent” Given: 4NH 3 + 5O 2  6H 2 O + 4NO 4 mol NO, works out exactly 4 mol NO, with leftover O 2 8 mol NO, with leftover O 2

7 In Chemistry: Q - How many moles of NO are produced if __ mol NH 3 are burned in __ mol O 2 ? (d) 4 mol NH 3, 2.5 mol O 2 Here, O 2 limits the production of NO; if there was more O 2, more NO would be produced Thus, O 2 is called the “limiting reagent” Sometimes the question is more complicated. For example, if grams of the two reactants are given instead of moles. Given: 4NH 3 + 5O 2  6H 2 O + 4NO 2 mol NO, leftover NH 3

8 8 Procedure for solving limiting reagent problems 1. If not already, balance the chemical equation. 2. Convert information given into units of moles. 3. Divide the mole amount of each reactant by its stoichiometric coefficient from the balanced chemical equation. 4. The reactant with the smallest quotient from step #3 will be the limiting reagent. 5. Work as typical stoichiometry problem using the limiting reagent.

9 9 Example #1 2Ca + O 2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? 1. If not already, balance the chemical equation. Balanced. 2. Convert information given into units of moles. Already in units of moles. 3. Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 15 moles Ca = moles O 2 = 12 2 moles Ca 1 mol O 2

10 10 Example #1 (con’t) 2Ca + O 2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? 4. The reactant with the smallest quotient from step #3 will be the limiting reagent. Calcium will be the limiting reagent in this problem. 5. Work as typical stoichiometry problem using the limiting reagent. 15mol Ca x 2 mol CaO = 15mol CaO 2 mol Ca 10

11 11 Example #2 What mass of CO 2 can be produced by the reaction of 8.0 grams of CH 4 with 48 grams of O 2 according to the equation below? CH 4 + 2O 2 → CO 2 + 2H 2 O 1. If not already, balance the chemical equation. Balanced. 2. Convert information given into units of moles. Information given is in units of grams. We need to use the molar mass of each reactant to find moles of each reactant. 8.0g CH 4 x 1mol CH 4 = 0.50mol CH g CH 4 48g O 2 x 1mol O 2 = 1.50mol O g O 2 11

12 12 Example #2 (con’t) What mass of CO 2 can be produced by the reaction of 8.0 grams of CH 4 with 48 grams of O 2 according to the equation below? CH 4 + 2O 2 → CO 2 + 2H 2 O 3. Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 0.5mol CH 4 = mol O 2 = mol CH 4 2mol O 2 4. The reactant with the smallest quotient from step #3 will be the limiting reagent. CH 4

13 13 Example #2 (con’t) What mass of CO 2 can be produced by the reaction of 8.0 grams of CH 4 with 48 grams of O 2 according to the equation below? CH 4 + 2O 2 → CO 2 + 2H 2 O 4. Work as typical stoichiometry problem using the limiting reagent. 0.5mol CH 4 x 1mol CO 2 x 44g CO 2 = 22g CO 2 1mol CH 4 1mol CO 2

14 Limiting Reagents: shortcut Do two separate calculations using both given quantities. The smaller answer is correct. Ex3. How many g NO are produced if 20 g NH 3 is burned in 30 g O 2 ? 4NH 3 + 5O 2  6H 2 O+ 4NO 4 mol NO 5 mol O 2 x 30 g O g NO= 30.0 g NO 1 mol NO x 1 mol O g O 2 x 4 mol NO 4 mol NH 3 x # g NO= 20 g NH g NO= 30.0 g NO 1 mol NO x 1 mol NH g NH 3 x

15 Assignment 1.2Al + 6HCl  2AlCl 3 + 3H 2 If 25 g of aluminum was added to 90 g of HCl, what mass of H 2 will be produced (try this two ways – with the steps & using the shortcut)? 2.N 2 + 3H 2  2NH 3 : If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 3.What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O 2 ? 4.When C 3 H 8 burns in oxygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 5.How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

16 2 1 mol Al 27.0 g Al x # mol Al =25 g Al = mol 2mol # mol HCl =90 g HCl 1 mol HCl 36.5 g HCl x = mol 6mol HCl is limiting. 3 mol H 2 6 mol HCl x # g H 2 = 90 g HCl 2.0 g H 2 1 mol H 2 x 1 mol HCl 36.5 g HCl x = 2.47 g H 2 = =0.411

17 Question 2: shortcut 2Al + 6HCl  2AlCl 3 + 3H 2 If 25 g aluminum was added to 90 g HCl, what mass of H 2 will be produced? 3 mol H 2 2 mol Al x 25 g Al = 2.78 g H g H 2 1 mol H 2 x 1 mol Al 27.0 g Al x 3 mol H 2 6 mol HCl x 90 g HCl = 2.47 g H g H 2 1 mol H 2 x 1 mol HCl 36.5 g HCl x

18 Question 3: shortcut N 2 + 3H 2  2NH 3 If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 2 mol NH 3 1 mol N 2 x # g NH 3 = 20 g N 2 = 24.3 g H g NH 3 1 mol NH 3 x 1 mol N g N 2 x 2 mol NH 3 3 mol H 2 x # g NH 3 = 5.0 g H 2 = 28.3 g H g NH 3 1 mol NH 3 x 1 mol H g H 2 x N 2 is the limiting reagent

19 Question 4: shortcut 4Al + 3O 2  2 Al 2 O 3 What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O 2 ? 2 mol Al 2 O 3 4 mol Al x # g Al 2 O 3 = 10.0 g Al = 18.9 g Al 2 O g Al 2 O 3 1 mol H 2 x 1 mol Al 27.0 g Al x 2 mol Al 2 O 3 3 mol O 2 x # g Al 2 O 3 = 20.0 g O 2 = 42.5 g Al 2 O g Al 2 O 3 1 mol H 2 x 1 mol O g O 2 x

20 Question 5: shortcut C 3 H 8 + 5O 2  3CO 2 + 4H 2 O When C 3 H 8 burns in oxygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 3 mol CO 2 1 mol C 3 H 8 x # g CO 2 = 15.0 g C 3 H 8 = 45.0 g CO g CO 2 1 mol CO 2 x 1 mol C 3 H g C 3 H 8 x 3 mol CO 2 5 mol O 2 x # g CO 2 = 60.0 g O 2 = 49.5 g CO g CO 2 1 mol CO 2 x 1 mol O g O 2 x 5. Limiting reagent questions give values for two or more reagents (not just one)

21 34g NH 3 19g H 2 excess 23.0L H 2 25g Mg excess 164.3g Ag 3 PO g Na 3 PO Assignment Con’t 6. How many grams of NH 3 can be produced from the reaction of 28g of N 2 and 25g of H 2 ? 7. How much of the excess reagent in Problem 6 is left over? 8. What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl? 9. How much of the excess reagent in Problem 3 is left over? 10. Silver nitrate and sodium phosphate are reacted in equal amounts of 200g each. How many grams of silver phosphate are produced? 11. How much of the excess reagent in problem 5 is left?

22 Assignment 12. Iron reacts with sulfur to form iron(II) sulfide. You have 32.0g of sulfur and 100g of iron. Calculate the number of grams of iron(II) sulfide that will form. 13. Zinc reacts with sulfuric acid in a single replacement reaction. You have 40g of zinc and 57g of sulfuric acid. What volume of hydrogen gas will form? 14. Silver nitrate and sodium bromide will react in a double replacement reaction. You have 24g of silver nitrate and 39g of sodium bromide. How many grams of sodium nitrate will form? 15. When hydrochloric acid reacts with iron(II) sulfide, a double replacement reaction is to be expected. You have 67g of hydrochloric acid and 58 of iron(II) sulfide. How many grams of hydrogen sulfide will form? 16. When aluminum is heated in oxygen, aluminum oxide is formed. You have 384g of each reactant. How many moles of aluminum oxide will form? g FeS 13.0L H g NaNO g H 2 S 6.45mol Al 2 O 3

23 Limiting Reagents C.P. 1. A mixture of 49.1g of tin(II) nitrate and 81.2g of sodium chloride react. How many grams of tin(II) chloride are produced? How much excess is left over? 2. Silicon monocarbide, commonly known as carborundum, is prepared by heating silicon dioxide in the presence of carbon. How many grams of silicon monocarbide can be formed from 100.0g of graphite and 100.0g of silicon dioxide? (Clue: CO 2 is the other product) g SnCl g SiC

24 Limiting Reagents C.P. 1. Silver nitrate and sodium phosphate are reacted in equal amounts of 137g each. How many grams of silver phosphate are produced? 2. How much of the excess reagent in #1 is left? g SnCl g SiC

25 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl? How much of the excess reagent in Problem 3 is left over? 25

26 Check Point #2 A mixture of 49.1g of tin(II) nitrate and 81.2g of sodium chloride react. How many grams of tin(II) chloride are produced? How much excess is left over? 26


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