Lab “Oreo Lab” 1

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

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

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 CO2. The amount of wood limits the amount of products, therefor, it is the limiting reagent 4

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

b) 4 mol NH3, 20 mol O2 c) 8 mol NH3, 20 mol O2 In Chemistry: Given: 4NH3 + 5O2  6H2O + 4NO Q - How many moles of NO are produced if __ mol NH3 are burned in __ mol O2? a) 4 mol NH3, 5 mol O2 b) 4 mol NH3, 20 mol O2 c) 8 mol NH3, 20 mol O2 In all of these cases, NH3 limits the production of NO; if there was more NH3, more NO would be produced Thus, NH3 is called the “limiting reagent” 4 mol NO, works out exactly 4 mol NO, with leftover O2 8 mol NO, with leftover O2

In Chemistry: Given: 4NH3 + 5O2  6H2O + 4NO Q - How many moles of NO are produced if __ mol NH3 are burned in __ mol O2? (d) 4 mol NH3, 2.5 mol O2 Here, O2 limits the production of NO; if there was more O2, more NO would be produced Thus, O2 is called the “limiting reagent” Sometimes the question is more complicated. For example, if grams of the two reactants are given instead of moles. 2 mol NO, leftover NH3

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

Example #1 2Ca + O2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? If not already, balance the chemical equation. Balanced. Convert information given into units of moles. Already in units of moles. Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 15 moles Ca = 7.5 12 moles O2 = 12 2 moles Ca 1 mol O2

Example #1 (con’t) 2Ca + O2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? The reactant with the smallest quotient from step #3 will be the limiting reagent. Calcium will be the limiting reagent in this problem. Work as typical stoichiometry problem using the limiting reagent. 15mol Ca x 2 mol CaO = 15mol CaO 2 mol Ca 10

Example #2 What mass of CO2 can be produced by the reaction of 8.0 grams of CH4 with 48 grams of O2 according to the equation below? CH4 + 2O2 → CO2 + 2H2O If not already, balance the chemical equation. Balanced. 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 CH4 x 1mol CH4 = 0.50mol CH4 16.04g CH4 48g O2 x 1mol O2 = 1.50mol O2 32.0g O2 11

Example #2 (con’t) What mass of CO2 can be produced by the reaction of 8.0 grams of CH4 with 48 grams of O2 according to the equation below? CH4 + 2O2 → CO2 + 2H2O Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 0.5mol CH4 = 0.5 1.5mol O2 = 0.75 1 mol CH4 2mol O2 The reactant with the smallest quotient from step #3 will be the limiting reagent. CH4

Example #2 (con’t) What mass of CO2 can be produced by the reaction of 8.0 grams of CH4 with 48 grams of O2 according to the equation below? CH4 + 2O2 → CO2 + 2H2O Work as typical stoichiometry problem using the limiting reagent. 0.5mol CH4 x 1mol CO2 x 44g CO2 = 22g CO2 1mol CH4 1mol CO2

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 NH3 is burned in 30 g O2? 4NH3 + 5O2 6H2O+ 4NO # g NO= 1 mol NH3 17.0 g NH3 x 4 mol NO 4 mol NH3 x 30.0 g NO 1 mol NO x 20 g NH3 35.3 g NO = 1 mol O2 32.0 g O2 x 4 mol NO 5 mol O2 x 30.0 g NO 1 mol NO x 30 g O2 22.5 g NO =

Assignment 2Al + 6HCl  2AlCl3 + 3H2 If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced (try this two ways – with the steps & using the shortcut)? N2 + 3H2  2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

2 1 mol Al 27.0 g Al x # mol Al = 25 g Al = 0.926 mol 2mol = 0.463 1 mol HCl 36.5 g HCl x # mol HCl = 90 g HCl = 2.466 mol 6mol =0.411 HCl is limiting. # g H2 = 1 mol HCl 36.5 g HCl x 3 mol H2 6 mol HCl x 2.0 g H2 1 mol H2 x 90 g HCl = 2.47 g H2

Question 2: shortcut 2Al + 6HCl  2AlCl3 + 3H2 If 25 g aluminum was added to 90 g HCl, what mass of H2 will be produced? 1 mol Al 27.0 g Al x 3 mol H2 2 mol Al x 2.0 g H2 1 mol H2 x 25 g Al = 2.78 g H2 1 mol HCl 36.5 g HCl x 3 mol H2 6 mol HCl x 2.0 g H2 1 mol H2 x 90 g HCl = 2.47 g H2

N2 is the limiting reagent Question 3: shortcut N2 + 3H2  2NH3 If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? # g NH3= 1 mol N2 28.0 g N2 x 2 mol NH3 1 mol N2 x 17.0 g NH3 1 mol NH3 x 20 g N2 = 24.3 g H2 # g NH3 = 1 mol H2 2.0 g H2 x 2 mol NH3 3 mol H2 x 17.0 g NH3 1 mol NH3 x 5.0 g H2 = 28.3 g H2 N2 is the limiting reagent

Question 4: shortcut 4Al + 3O2  2 Al2O3 What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? # g Al2O3= 1 mol Al 27.0 g Al x 2 mol Al2O3 4 mol Al x 102.0 g Al2O3 1 mol H2 x 10.0 g Al = 18.9 g Al2O3 # g Al2O3= 1 mol O2 32.0 g O2 x 2 mol Al2O3 3 mol O2 x 102.0 g Al2O3 1 mol H2 x 20.0 g O2 = 42.5 g Al2O3

Question 5: shortcut C3H8 + 5O2  3CO2 + 4H2O When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? # g CO2= 1 mol C3H8 44.0 g C3H8 x 3 mol CO2 1 mol C3H8 x 44.0 g CO2 1 mol CO2 x 15.0 g C3H8 = 45.0 g CO2 # g CO2= 1 mol O2 32.0 g O2 x 3 mol CO2 5 mol O2 x 44.0 g CO2 1 mol CO2 x 60.0 g O2 = 49.5 g CO2 5. Limiting reagent questions give values for two or more reagents (not just one)

6.4 Assignment Con’t 34g NH3 19g H2 excess 23.0L H2 25g Mg excess How many grams of NH3 can be produced from the reaction of 28g of N2 and 25g of H2? How much of the excess reagent in Problem 6 is left over? 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? Silver nitrate and sodium phosphate are reacted in equal amounts of 200g each. How many grams of silver phosphate are produced? How much of the excess reagent in problem 5 is left? 34g NH3 19g H2 excess 23.0L H2 25g Mg excess 164.3g Ag3PO4 135.6g Na3PO4 21

6.4 Assignment 87.9g FeS 13.0L H2 11.9g NaNO3 22.5g H2S 6.45mol Al2O3 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. 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? 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? 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? When aluminum is heated in oxygen, aluminum oxide is formed. You have 384g of each reactant. How many moles of aluminum oxide will form? 87.9g FeS 13.0L H2 11.9g NaNO3 22.5g H2S 6.45mol Al2O3 22

6.4 Limiting Reagents C.P. 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? 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: CO2 is the other product) 38.3g SnCl2 66.73g SiC 23

6.4 Limiting Reagents C.P. Silver nitrate and sodium phosphate are reacted in equal amounts of 137g each. How many grams of silver phosphate are produced? How much of the excess reagent in #1 is left? 38.3g SnCl2 66.73g SiC 24

What volume of hydrogen at STP is produced from the reaction of 50 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?

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?