Sample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O


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1 STOICHIOMETRY and percent yield calculations 1 Steps for solving Stoichiometric Problems 2 Step 1 Write the balanced equation for the reaction. Step 2 Identify your known and unknown quantities. Step 3 Convert from grams of known to moles of known Step 4 Convert from moles of known to moles of unknown Step 5 Convert from moles of unknown to grams of unknown Reactants: Zn + I 2 Product: Zn I 2 Sample Problem: How much H 2 O will be formed if 454 g of NH 4 NO 3 decomposes? 3 SAMPLE PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much H 2 O is formed? STEP 1 Write the balanced chemical equation 4 SAMPLE PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much H 2 O is formed? STEP 2 Identify your known and unknown quantities. 5 STEP 3 Convert from grams of known to moles of known Use the molar mass 454 g 1 mol g = 5.68 mol NH 4NO3 6
2 NH 4 NO 3 > > N 2 O + 2 H 2 O STEP 4 Convert from moles of known to moles of unknown Use the mole ratio 1 mol NH 4 NO 3 : 2 mol H 2 O 2 mol H 2 O produced 1 mol NH 4 NO 3 used 7 NH 4 NO 3 > > N 2 O + 2 H 2 O STEP 4 Convert from moles of known to moles of unknown Use the mole ratio 5.68 mol NH 4 NO 3 2 mol H 2 O produced 1 mol NH 4 NO 3 used = 11.4 mol H 2 O produced 8 STEP 5 Convert from moles of unknown to grams of unknown Use the molar mass 11.4 mol H 2 O g 1 mol = 204 g H 2 O 9 Molar known GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS known known Mole Ratio unknown Molar unknown unknown 10 ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS! SAMPLE PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much H 2 O is formed? OUR FINAL ANSWER WAS: = 204 g H 2 O produced This is called the THEORETICAL YIELD it is how much water we theoretically should produce. 11 Theoretical Yield: a.k.a. predicted yield is calculated (by stoichometry) The amount of product we should get. Actual Yield: aka Experimental Yield is measured in the lab What you actually did produce Always less than the theoretical yield. 12
3 Percent Yield = Actual Yield Theoretical Yield Always less than 100 % Ratio of actual production to theoretical production. Not the same as percent error! x NH 4 NO 3 > > N 2 O + 2 H 2 O Sample Problem Con t: Our theoretical yield was 204 g H 2 O If Johnny Q. Chemistry did the reaction and only collected 186 g H 2 O, the percent yield would be: 186 g H 2 O 204 g H 2 O 14 x 100 = 91% Stoichiometry can be used to Determine a Formula: 15 Determining the Formula of a Hydrocarbon by Combustion 16 If we burn an hydrocarbon fuel with an unknown formula, C x H y,, using a known quantity of oxygen gas: 4.0 g C x H y g O 2 > 5.0 g CO g H 2 O CCR, page 138 The mass of CO 2 and H 2 O produced can be used to determine the amount of C and H present in the formula. Reactions Involving a LIMITING REACTANT Usually, there is not enough of one reagent to use up the other reagent completely. LIMITS the The reagent in short supply LIMITS the quantity of product that can be formed. It is called the limiting reagent 17 LIMITING REACTANTS (See CD Screen 4.8) React solid Zn with mol HCl (aq) Zn + 2 HCl > > ZnCl 2 + H Rxn 1: Balloon inflates fully, some Zn left * More than enough Zn to use up the mol HCl Rxn 2: Balloon inflates fully, no Zn left * Right amount of each (HCl( and Zn) Rxn 3: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up mol HCl 18
4 LIMITING REACTANTS React solid Zn with mol HCl (aq) Zn + 2 HCl > > ZnCl 2 + H 2 19 LIMITING REACTANTS 20 Rxn 1 Rxn 2 Rxn 3 mass Zn (g) ~ mol Zn mol HCl mol HCl/mol Zn 1.00/1 2.00/1 5.00/1 Lim Reactant LR = HCl no LR LR = Zn Demo of limiting reactants on Screen g of NH 4 NO 3 > > N 2 O + 2 H 2 O STEP 6 Calculate the percent yield % yield = % yield = 131 g 250. g actual yield theoretical yield 100% 100% = 52.4% 21 Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of Matter Demo of conservation of matter, See Screen LIMITING REACTANTS 23 Reaction to be Studied 24 Reactantseactants Products 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = Excess reactant =
5 PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? reactant product reactant Stoichiometric factor product Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. 27 Deciding on the Limiting Reactant 28 If mol Cl 2 mol Al > 3 2 Reactants must be in the mole ratio mol Cl 2 = 3 mol Al 2 There is not enough Al to use up all the Cl 2 Lim reag = Al Deciding on the Limiting Reactant If mol Cl 2 mol Al < 3 2 There is not enough Cl 2 to use up all the Al Lim reag = Cl 2 29 Step 2 of LR problem: Calculate moles of each reactant We have 5.40 g of Al and 8.10 g of Cl g Al 1 mol 27.0 g = mol Al 8.10 g Cl 2 1 mol 70.9 g = mol Cl 2 30
6 Find mole ratio of reactants mol Cl 2 mol Al = mol mol = 0.57 This would be 3/2, or 1.5/1, if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl 2 31 Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Limiting reactant = Cl 2 Base all calcs.. on Cl 2 grams Cl 2 moles Cl 2 1 mol Al 2 Cl 6 3 mol Cl 2 grams Al 2 Cl 6 moles Al 2 Cl 6 32 CALCULATIONS: calculate mass of Al 2 Cl 6 expected. Step 1: Calculate moles of Al 2 Cl 6 expected based on LR mol Cl 2 1 mol Al 2Cl 6 3 mol Cl 2 = mol Al 2 Cl 6 Step 2: Calculate mass of Al 2 Cl 6 expected based on LR mol Al 2 Cl g Al 2Cl 6 mol = 10.1 g Al 2 Cl 6 33 How much of which reactant will remain when reaction is complete? Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required. Then find how much Al is in excess. 34 Calculating Excess Al 2 Al + 3 Cl 2 products mol mol = LR mol Cl 2 2 mol Al 3 mol Cl 2 = mol Al req' d Excess Al = Al available  Al required = mol mol = mol Al in excess 35 Using Stoichiometry to Determine a Formula C x H y + some oxygen > g CO g H 2 O First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. +O g CO 2 Puddle of C x H y g 1 CO 2 molecule forms for each C atom in C x H y +O g H 2 O 1 H 2 O molecule forms for each 2 H atoms in C x H y 36
7 Using Stoichiometry to Determine a Formula C x H y + some oxygen > g CO g H 2 O First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1. Calculate amount of C in CO x 103 mol CO 2 > > 8.61 x 103 mol C 2. Calculate amount of H in H 2 O x 103 mol H 2 O  >1.149 x 102 mol H 37 Using Stoichiometry to Determine a Formula C x H y + some oxygen > g CO g H 2 O Now find ratio of mol H/mol C to find values of x and y in C x H y x 102 mol H/ H 8.61 x 103 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3 H 4 38
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