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Mullis1 Stoichiometry (S) Composition S: Mass relationships in compounds Reaction S: Mass relationships between reactants and products To find amounts.

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Presentation on theme: "Mullis1 Stoichiometry (S) Composition S: Mass relationships in compounds Reaction S: Mass relationships between reactants and products To find amounts."— Presentation transcript:

1 Mullis1 Stoichiometry (S) Composition S: Mass relationships in compounds Reaction S: Mass relationships between reactants and products To find amounts of products and/or reactants, you must convert both to moles!

2 Mullis2 Mole Ratio A mole ratio is a conversion factor that relates the amounts of moles of any 2 substances involved in a chemical reaction. 2Al 2 O 3 (l)4Al(s) + 3O 2 (g) Possible mole ratios: 2 mol Al 2 O 3or 4 mol Al 4 mol Al 2 mol Al 2 O 3 2 mol Al 2 O 3or 3 mol O 2 3 mol O 2 2 mol Al 2 O 3 4 mol Al or 3 mol O 2 3 mol O 2 4 mol Al

3 Mullis3 Ideal Condition In chemical reactions, all reactants are converted to products under ideal conditions. Solution plan if you have a given and unknown quantity in moles: Given mol x Mole ratio unknown mol = Unknown mol given mol

4 Mullis4 Given mol x Mole ratio unknown mol = Unknown mol given mol 2Al 2 O 3 (l)4Al(s) + 3O 2 (g) If start with 3 moles aluminum oxide, how much Al will be obtained? 3 mol Al 2 O 3 4 mol Al = 6 mol Al 2 mol Al 2 O 3 6 mol Al26.98 g Al= 161.88 g Al 1 mol Al

5 Mullis5 Conversion Reminders

6 Mullis6 4 Problem Types 1 mole O 2 2 moles H 2 O X moles H 2 O 1moles A  moles B2 steps 1moles A  grams B3 steps 32 g O 2 1 mole O 2 2 moles H 2 O X moles H 2 O 1grams A  moles B3 steps 1 mole O 2 2 moles H 2 1 mole H 2 2.02 g H 2 X grams H 2 1grams A  grams B4 steps 32 g O 2 1 mole O 2 2 mole H 2 1 mole H 2 2.02 g H 2 X grams H 2

7 Mullis7 Practice Problems 1. When sodium azide (NaN 3 ) is activated in an automobile airbag, nitrogen gas and sodium are produced. If 0.500 mol NaN 3 react, what mass in grams of nitrogen would result? 2. Carborundum, SiC, is a hard substance made by combining silicon dioxide with coke (C). The products are SiC and CO. What is the mass of SiC in grams from the complete reaction of 2.00 mol carbon?

8 Mullis8 Practice Problems: Stoichiometry Set 4 1.How many grams of sodium should be added to 50 g fluorine to make sodium fluoride? 2.Mg(OH) 2 + 2HCl2H 2 O + MgCl 2 This equation represents the neutralization of stomach acid by milk of magnesia. a.What is the mass in grams of MgCl 2 which will be produced if 3.00 g of Mg(OH) 2 reacts? b.What mass in grams of HCl is required to completely react with 3.00 g of Mg(OH) 2 ?

9 Mullis9 % Yield % Yield = Actual – Expected x 100% Expected Which is the same as: % Yield = Observed–Theoretical x 100% Theoretical To find Expected, or Theoretical, yield: Do a gram-to-gram problem.

10 Mullis10 % Yield Practice 1.For the reaction SO 3 + H 2 O  H 2 SO 4, calculate the percent yield if 500. g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid. 2.For the reaction 2KClO 3  2KCl + 3O 2, calculate the percent yield if 10. g of potassium chlorate is heated to produce 2.5 g of oxygen. 3.For the reaction 2KClO 3  2KCl + 3O 2, calculate the percent yield if 10. g of potassium chlorate is heated to produce 3.0 g of potassium chlorate. 1.93.9% 2.65 % 3.49%

11 Mullis11 % Yield Practice 2 1.For the reaction SO 3 + H 2 O  H 2 SO 4, calculate the percent yield if 500. g of sulfur trioxide react with excess water to produce 475 g of sulfuric acid. 2.For the reaction 2HCl + Mg(OH) 2  MgCl 2 + 2H 2 O, calculate the percent yield if 3.65 g of HCl reacts to produce 3.0 g of magnesium chloride. 3.For the reaction 2HCl + Mg(OH) 2  MgCl 2 + 2H 2 O, calculate the percent yield if 3.65 g of HCl reacts to produce 4.0 g of magnesium chloride. 1.77.6% 2.63% 3.84%

12 Mullis12 Limiting Reactants A limiting reactant is the one that limits the amount of product that can be obtained. An excess reactant is one that is not used up completely in a reaction. 6 bottles 8 corks Limiting reactant: bottles Excess reactant: corks

13 Mullis13 Sample Problem Zinc citrate, Zn 3 (C 6 H 5 O 7 ) 2 is found in some toothpastes. It is synthesized by the reaction of zinc carbonate with citric acid. 3ZnCO 3 +2C 6 H 8 O 7 Zn 3 (C 6 H 5 O 7 ) 2 +3H 2 O + CO 2 With 1 mole each of ZnCO 3 and C 6 H 8 O 7, which is the limiting reactant? 1 mol ZnCO 3 x 2mole C 6 H 8 O 7 =0.67 mol C 6 H 8 O 7 needed 3 mole ZnCO 3  1 mol C 6 H 8 O 7 x 3 mole ZnCO 3 = 1.5 mol ZnCO 3 needed 2mole C 6 H 8 O 7

14 Mullis14 Sample Problem, cont. 3ZnCO 3 +2C 6 H 8 O 7 Zn 3 (C 6 H 5 O 7 ) 2 +3H 2 O + CO 2 If there is 6 mol ZnCO 3 and 10 mol C 6 H 8 O 7, which is the reactant in excess? 6 mol ZnCO 3 x 2mole C 6 H 8 O 7 = 4 mol C 6 H 8 O 7 needed 3 mole ZnCO 3  10 mol C 6 H 8 O 7 x 3 mole ZnCO 3 = 15 mol ZnCO 3 needed 2mole C 6 H 8 O 7 C 6 H 8 O 7 is in excess.

15 Mullis15 Limiting Reactants, Sample Problems 1 1.Benzene and chlorine react to form chlorobenzene and HCl: C 6 H 6 (l) + Cl 2 (g)C 6 H 5 Cl(s) + HCl(g) a.What is the limiting reactant if 2 moles C 6 H 6 are mixed with 1 mole Cl 2 ? b.What is the limiting reactant if 0.5 moles C 6 H 6 are mixed with 0.75 moles Cl 2 ? c.If Cl 2 is provided in excess, how much C 6 H 5 Cl is formed when 36.8 g of C 6 H 6 is added?

16 Mullis16 Limiting reactants using mass Ca 3 (PO 4 ) 2 + 3H 2 SO 4 3CaSO 4 + 2H 3 PO 4 If 250 g of Ca 3 (PO 4 ) 2 react with 3 mol of H 2 SO 4, will 3 mol of CaSO 4 be formed? 250 g Ca 3 (PO 4 ) 2 x 1 mol = 0.81 mol Ca 3 (PO 4 ) 2 310.18 g Molar mass Ca 3 (PO 4 ) 2 = (3 x 40.08) + (2 x 30.97)+ (8 x 16) = 310.18 g/mol 0.81 mol Ca 3 (PO 4 ) 2 x 3 mol CaSO 4 = 2.43 mol CaSO 4 1 mol Ca 3 (PO 4 ) 2 No- we have only 2.43 mol CaSO 4.

17 Mullis17 Finding remaining Amounts of excess Balanced equations can only be used to calculate the amount of substances that react. To find how much is left over, or how much DOES NOT react, you must take the difference between starting and the amount that reacts. 8 Zn(s) + S 8 (s)8 ZnS(s) If 2.00 mol Zn are heated with 1.00 mol S 8, which is the limiting reactant? 2 mol Zn x 1 mol S 8 = 0.25 mol S 8 1 mol S 8 x 8 mol Zn = 8 mol Zn 8 mol Zn 1 mol S 8 How many moles of excess reactant remain? 1.00 mol S 8 - 0.25 mol S 8 = 0.75 mol S 8

18 Mullis18 Limiting Reactants, Sample Problems 2 1.Methanol, CH 3 OH, is synthesized by the reaction of hydrogen and carbon monoxide. a. If 500 mol of CO and 750 mol of H 2 are present, which is the limiting reactant? b. How many moles of the excess reactant remain unchanged? c. How moles of CH 3 OH are formed? 2. In the formation of zinc citrate, how many moles of Zn 3 (C 6 H 5 O 7 ) 2 would be produced with 6 mol ZnCO 3 and 10 mol C 6 H 8 O 7 ? See your notes and use the equation and the limiting reactant to solve.


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