1Chapter 8 Stoichiometry: Moles meet Reactions! Calculating quantities from chemical reactions.
2Write the equation for combination reaction between sodium and chlorine. Na Cl NaCl22The above equation can mean:1. 2 Na atoms + 1 Cl2 molecule yield2 NaCl formula units.2. 2 doz. Na atoms + 1 doz. Cl2 molecules yield2 doz. NaCl formula units.3. 2 mol Na + 1 mol Cl2 yields 2 mol NaClMole Ratio!!The COEFFICIENTS in the balanced equation tell ____________________________________________________therelative number of moles of reactants and products.
3Mol – Mol Conversions using Mole Ratio From the coefficients in the balanced equation we can write mole ratios:2 mol Na = 1 mol Cl22 mol Na = 2 mol NaCl (equals 1:1 ratio)1 mol Cl2 = 2 mol NaClThese mole ratios can be used as conversion factors to solve problems.
4mole ratio = 1mol Cl2: 2mol NaCl Problem - 1We can now use the appropriate mole ratio to solve each problem:Ex. 1 How many moles of NaCl will form from 2.7 moles of Cl2 ?mole ratio = 1mol Cl2: 2mol NaCl2 mol NaCl2.7 mol Cl2x =5.4mol NaCl1 mol Cl2G:M:M:G
5mole ratio = 2 mol Na : 1 mol Cl2 Problem - 2How many moles of Na are needed to react with 4.5 mol of Cl2?mole ratio = 2 mol Na : 1 mol Cl22 mol Na4.5 mol Cl29.0mol Nax =1 mol Cl2G:M:M:G
6Mole- Gram Conversions We already know that 1 mol = gfm2Na + 1Cl2 2 NaClCalculate the gfm’s:1 mol Na = ________ g231(23)1 mol Cl2 = ________ g702(35)1 mol NaCl = ________ g581(23) + 1(35)
7Now we will use BOTH the mole ratios and the gfms to solve problems. 1234
81. How many moles of Cl2 are needed to form 50.0 g of NaCl? mol NaClmol Cl2150.0 g NaClxx= mol Cl2g NaCl2mol NaCl58= .431 mol Cl2G:M:M:G
92. How many grams of NaCl are formed from 3.1 moles of Na? mol NaClg NaCl583.1 mol Naxx= g NaCl2mol Na1mol NaCl= 180 g NaClG:M:M:G
10How many grams of Na are needed to react with 24.5g of Cl2? 1mol Cl2g Na2mol Na2324.5 g Cl2xxx1mol Cl2g Cl21mol Na70= g Na16.1G:M:M:G
114. How many grams of Cl2 are needed to form 44.0 g of NaCl? 1mol NaCl1mol Cl270g Cl244.0 g NaClxxx2mol NaCl58g NaCl1mol Cl2= g Cl226.6G:M:M:G
12Limiting Reagent Problems Note - A reagent is the same as a reactant.the reactant that limits theamount of product formed.Limiting Reagent -The Limiting Reagent is the reactant that determines how much product can be produced. Because you run out of this reactant no more product is made.
13Limiting Reagent Example 1: When 10.0 grams of hydrogen combines with 20.0 grams of oxygen, how many grams of water will be produced?Write the balanced equation:2H2 + O2 2H2OH O N Cl Br I FCalculate the gfms:H2 =O2 =H2O =2(1)2(16)2(1) + 1(16)= 2 g= 32 g= 18 g
14G:M:M:G = g H2O = g H2O 22.5 mol H2 1 2 mol H2O 18 g H2O 10.0gH2 2 We don’t know which reactant will limit the amount of water that is formed, so we must figure out the amount of product for each of the given amounts of reactants.In essence we perform 2 separate problems.mol H212mol H2O18g H2O10.0gH2xxx2g H22mol H21mol H2O= g H2O90.01mol O22mol H2O18g H2O20 g O2xxx32g O21mol O21mol H2O= g H2O22.5G:M:M:G
15When the two given amounts of reactants are combined in a chemical reaction, the lesseramount calculated above will be the amount thatactually forms. The reactant that yields theLEAST amount of product is called the “limitingreagent.”The amount of product formed in the above reaction:22.5 gTheoretical YieldThe limiting reagent in the above reaction is:OxygenLets say students performed this reaction and they obtained 20.2 g of water. This is the Actual Yield.
16% yieldis the ratio of actual yield (lab) to the theoretical yield (calculated)% yield =Actual / Theoretical x 100% yield = / X 100 == 89.78%
172Na + 1Cl2 2 NaCl G:M:M:G = g NaCl = g NaCl mol Cl2 1 2 mol NaCl 58 If 20.0 g of chlorine is combined with 20.0 g ofsodium, how many g of NaCl will form?In essence we perform 2 separate problems.mol Cl212mol NaCl58g NaCl20.0 gCl2xxx70g Cl21mol Cl21mol NaClLimiting Reagent= g NaCl33.1Theoretical Yield1mol Na2mol NaCl58g NaCl20.0 g Naxxx23g Na2mol Na1mol NaCl= g NaCl50.42Na + 1Cl2 2 NaClG:M:M:G
18While creating NaCl in the lab, 27. 8g on salt were produced While creating NaCl in the lab, 27.8g on salt were produced. What is the % yield of salt?% Yield = Actual / Theoretical x 100% yield = / 33.1 x 100 == 69.09%