Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 8 Stoichiometry: Moles meet Reactions! Calculating quantities from chemical reactions.

Similar presentations


Presentation on theme: "Chapter 8 Stoichiometry: Moles meet Reactions! Calculating quantities from chemical reactions."— Presentation transcript:

1 Chapter 8 Stoichiometry: Moles meet Reactions! Calculating quantities from chemical reactions.

2 Write the equation for combination reaction between sodium and chlorine. Na + Cl 2 NaCl2 The above equation can mean: 1. 2 Na atoms + 1 Cl 2 molecule yield 2 NaCl formula units doz. Na atoms + 1 doz. Cl 2 molecules yield 2 doz. NaCl formula units mol Na + 1 mol Cl 2 yields 2 mol NaCl Mole Ratio!! The COEFFICIENTS in the balanced equation tell ______ ______________________________________________ the relative number of moles of reactants and products. 2

3 Mol – Mol Conversions using Mole Ratio From the coefficients in the balanced equation we can write mole ratios: 2 mol Na = 1 mol Cl 2 2 mol Na = 2 mol NaCl (equals 1:1 ratio) 1 mol Cl 2 = 2 mol NaCl These mole ratios can be used as conversion factors to solve problems.

4 Problem - 1 We 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 Cl 2 ? mole ratio = 1mol Cl 2 : 2mol NaCl 2.7 mol Cl 2 mol NaCl x = 1 mol Cl 2 2 mol NaCl 5.4 G:M:M:G

5 Problem - 2 How many moles of Na are needed to react with 4.5 mol of Cl 2 ? mole ratio = 2 mol Na : 1 mol Cl mol Cl 2 mol Na x = 1 mol Cl 2 2 mol Na 9.0 G:M:M:G

6 Mole- Gram Conversions We already know that 1 mol = gfm  2Na + 1Cl 2  2 NaCl Calculate the gfm’s: 1 mol N a = ________ g 1(23) 23 1 mol Cl 2 = ________ g 2(35) 70 1 mol NaCl = ________ g 1(23) + 1(35) 58

7 Problems Now we will use BOTH the mole ratios and the gfms to solve problems. 1234

8 1. How many moles of Cl 2 are needed to form 50.0 g of NaCl? 50.0 g NaCl x g NaCl mol NaCl 58 1 x mol NaCl mol Cl = mol Cl 2 =.431 mol Cl 2 G:M:M:G

9 2. How many grams of NaCl are formed from 3.1 moles of Na? 3.1 mol Na x mol Na mol NaCl 2 2 x mol NaCl g NaCl 1 58 = g NaCl = 180 g NaCl G:M:M:G

10 3.How many grams of Na are needed to react with 24.5g of Cl 2 ? 24.5 g Cl 2 x g Cl 2 mol Cl x mol Cl 2 mol Na 1 2 = g Na x mol Na1 23 g Na G:M:M:G 16.1

11 4. How many grams of Cl 2 are needed to form 44.0 g of NaCl? 44.0 g NaCl x g NaCl mol NaCl 58 1 x mol NaCl mol Cl = g Cl 2 x mol Cl g Cl G:M:M:G

12 Limiting Reagent Problems Note - A reagent is the same as a reactant. Limiting Reagent - the reactant that limits the amount of product formed. 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.

13 Write the balanced equation: 2H 2 + O 2  2H 2 OH O N Cl Br I F Calculate the gfms: H 2 = O 2 = H 2 O = 2(1) 2(16) 2(1) + 1(16) = 2 g = 32 g = 18 g Limiting Reagent Example 1: When 10.0 grams of hydrogen combines with 20.0 grams of oxygen, how many grams of water will be produced?

14 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. 10.0gH 2 x x g H 2 mol H mol H 2 O 2 2 g H 2 O 1 18 = g H 2 O 20 g O 2 x g O 2 mol O x mol O 2 mol H 2 O 1 2 x mol H 2 O g H 2 O 1 18 = g H 2 O G:M:M:G

15 When the two given amounts of reactants are combined in a chemical reaction, the lesser amount calculated above will be the amount that actually forms. The reactant that yields the LEAST amount of product is called the “limiting reagent.” The amount of product formed in the above reaction: 22.5 g The limiting reagent in the above reaction is: Oxygen Theoretical Yield Lets say students performed this reaction and they obtained 20.2 g of water. This is the Actual Yield.

16 % yield is the ratio of actual yield (lab) to the theoretical yield (calculated) % yield = Actual / Theoretical x 100 % yield = 20.2 / 22.5 X 100 = = 89.78%

17 In essence we perform 2 separate problems gCl 2 x x g Cl 2 mol Cl mol Cl 2 mol NaCl 1 2 g NaCl 1 58 = g NaCl 20.0 g Na x g Na mol Na 23 1 x mol Na mol NaCl 2 2 x mol NaCl g NaCl 1 58 = g NaCl If 20.0 g of chlorine is combined with 20.0 g of sodium, how many g of NaCl will form? 2Na + 1Cl 2  2 NaCl Limiting Reagent G:M:M:G Theoretical Yield

18 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%


Download ppt "Chapter 8 Stoichiometry: Moles meet Reactions! Calculating quantities from chemical reactions."

Similar presentations


Ads by Google