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**Chapter 8 Stoichiometry: Moles meet Reactions!**

Calculating quantities from chemical reactions.

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**Write the equation for combination reaction between sodium and chlorine.**

Na Cl NaCl 2 2 The above equation can mean: 1. 2 Na atoms + 1 Cl2 molecule yield 2 NaCl formula units. 2. 2 doz. Na atoms + 1 doz. Cl2 molecules yield 2 doz. NaCl formula units. 3. 2 mol Na + 1 mol Cl2 yields 2 mol NaCl Mole Ratio!! The COEFFICIENTS in the balanced equation tell ______ ______________________________________________ the relative number of moles of reactants and products.

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**Mol – Mol Conversions using Mole Ratio**

From the coefficients in the balanced equation we can write mole ratios: 2 mol Na = 1 mol Cl2 2 mol Na = 2 mol NaCl (equals 1:1 ratio) 1 mol Cl2 = 2 mol NaCl These mole ratios can be used as conversion factors to solve problems.

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**mole ratio = 1mol Cl2: 2mol NaCl**

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 Cl2 ? mole ratio = 1mol Cl2: 2mol NaCl 2 mol NaCl 2.7 mol Cl2 x = 5.4 mol NaCl 1 mol Cl2 G:M:M:G

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**mole ratio = 2 mol Na : 1 mol Cl2**

Problem - 2 How many moles of Na are needed to react with 4.5 mol of Cl2? mole ratio = 2 mol Na : 1 mol Cl2 2 mol Na 4.5 mol Cl2 9.0 mol Na x = 1 mol Cl2 G:M:M:G

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**Mole- Gram Conversions**

We already know that 1 mol = gfm 2Na + 1Cl2 2 NaCl Calculate the gfm’s: 1 mol Na = ________ g 23 1(23) 1 mol Cl2 = ________ g 70 2(35) 1 mol NaCl = ________ g 58 1(23) + 1(35)

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**Now we will use BOTH the mole ratios and the gfms to solve problems.**

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**1. How many moles of Cl2 are needed to form 50.0 g of NaCl?**

mol NaCl mol Cl2 1 50.0 g NaCl x x = mol Cl2 g NaCl 2 mol NaCl 58 = .431 mol Cl2 G:M:M:G

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**2. How many grams of NaCl are formed from 3.1 moles of Na?**

mol NaCl g NaCl 58 3.1 mol Na x x = g NaCl 2 mol Na 1 mol NaCl = 180 g NaCl G:M:M:G

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**How many grams of Na are needed to react with 24.5g of Cl2?**

1 mol Cl2 g Na 2 mol Na 23 24.5 g Cl2 x x x 1 mol Cl2 g Cl2 1 mol Na 70 = g Na 16.1 G:M:M:G

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**4. How many grams of Cl2 are needed to form 44.0 g of NaCl?**

1 mol NaCl 1 mol Cl2 70 g Cl2 44.0 g NaCl x x x 2 mol NaCl 58 g NaCl 1 mol Cl2 = g Cl2 26.6 G:M:M:G

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**Limiting Reagent Problems**

Note - A reagent is the same as a reactant. the reactant that limits the amount 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.

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**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? Write the balanced equation: 2H2 + O2 2H2O H O N Cl Br I F Calculate the gfms: H2 = O2 = H2O = 2(1) 2(16) 2(1) + 1(16) = 2 g = 32 g = 18 g

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**G: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 H2 1 2 mol H2O 18 g H2O 10.0gH2 x x x 2 g H2 2 mol H2 1 mol H2O = g H2O 90.0 1 mol O2 2 mol H2O 18 g H2O 20 g O2 x x x 32 g O2 1 mol O2 1 mol H2O = g H2O 22.5 G:M:M:G

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**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 Theoretical Yield The limiting reagent in the above reaction is: Oxygen Lets say students performed this reaction and they obtained 20.2 g of water. This is the Actual Yield.

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

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**2Na + 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 of sodium, how many g of NaCl will form? In essence we perform 2 separate problems. mol Cl2 1 2 mol NaCl 58 g NaCl 20.0 gCl2 x x x 70 g Cl2 1 mol Cl2 1 mol NaCl Limiting Reagent = g NaCl 33.1 Theoretical Yield 1 mol Na 2 mol NaCl 58 g NaCl 20.0 g Na x x x 23 g Na 2 mol Na 1 mol NaCl = g NaCl 50.4 2Na + 1Cl2 2 NaCl G:M:M:G

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**While 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%

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Unit 11 Stoichiometry CP Chemistry.

Unit 11 Stoichiometry CP Chemistry.

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