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Lecturer: Amal Abu- Mostafa.  Available Ingredients ◦ 4 slices of bread ◦ 1 jar of peanut butter ◦ 1/2 jar of jelly Limiting Reactant Limiting Reactant.

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Presentation on theme: "Lecturer: Amal Abu- Mostafa.  Available Ingredients ◦ 4 slices of bread ◦ 1 jar of peanut butter ◦ 1/2 jar of jelly Limiting Reactant Limiting Reactant."— Presentation transcript:

1 Lecturer: Amal Abu- Mostafa

2  Available Ingredients ◦ 4 slices of bread ◦ 1 jar of peanut butter ◦ 1/2 jar of jelly Limiting Reactant Limiting Reactant Bread Excess Reactants Excess Reactants peanut butter and jelly

3  The limiting reactant (or limiting reagent): is the reactant that is entirely consumed (used up)when a reaction goes to completion.  Excess reactant is: a reactant that is not completely consumed in the reaction.  Once one of the reactants is used up, the reaction stops.  This means that: the moles of product are always determined by the starting moles of limiting reactant.

4  Zinc metal reacts with hydrochloric acid by the following reaction:  Zn (s) + 2HCl (aq)  ZnCl 2(aq) + H 2(g)  If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, which reactant is the limiting reactant? And how many moles of H 2 are produced?

5  Step 1: Which is the limiting reactant?  The reactant that gives the smaller amount of product is the limiting reactant.  Step 2: You obtain the amount of product actually obtained from the amount of limiting reactant.

6  Zn (s) + 2 HCl (aq)  ZnCl 2(aq) + H 2(g)  Step 1: From the equation: First:  1 mol of Zn  1 mol of H 2  0.30 mol of Zn  ?? Mol of H 2  Moles of H 2 =0.3× 1 = 0.3 mol 1  Second:  2 mol of HCl  1 mol of H 2  0.52 mol of HCl  ?? of H 2  Moles of H 2 = 0.52 × 1= 0.26 mol 2

7  So hydrochloric acid HCl must be the limiting reactant because it gave the smaller amount of the product which is H 2.  (Zinc is the excess reactant.)  Step 2: Since HCl is the limiting reactant, the amount of H 2 produced = 0.26 mol.

8  The theoretical yield of product:  Is the maximum amount of product that can be obtained by a reaction from given amounts of reactants (in the balanced equation).

9  Is the amount of products actually obtained when the reaction takes place.  The actual yield of a product may be much less than the theoretical yield for several possible reasons.  It is important to realize that the actual yield is an experimentally determined quantity.

10  The percentage yield of product is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). measured in lab  Percent yield = Actual yield (g) x 100 Theoretical yield (g) calculated on paper

11  When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and percentage % yields of KCl.  K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2  45.8 g ?? g  Solution:  The actual yield of KCl = 46.3 g  First we should calculate:  Moles of K 2 CO 3 = m = 45.8 = 0.33 mol  M 138.2

12  From the equation  K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2  1mol of K 2 CO 3  2 mol of KCl  0.33 mol of K 2 CO 3  ?? mol of KCl  Moles of KCl = 0.33 × 2 = 0.66 mol  1  Theoretical yield of KCl (m) = n × M  = 0.66 ×  = 49.2 g  Actual yield of KCl = 46.3 g  Percent yield = Actual yield (g) x 100 Theoretical yield (g)  = 46.3 × 100 = 94.1% 49.2

13  In the reaction: 4 Al + 3 O 2 → 2 Al 2 O 3  A) calculate the theoretical yield of Al 2 O 3 if 54 g of Al is reacted with enough O 2  B) if the experimental yield was 51 g calculate the percentage yield

14  A)Moles of Al = m = 54 = 2 mol M 27  From the equation:  4 Al + 3 O 2 → 2 Al 2 O 3  4 mol of Al → 2 mol of Al 2 O 3  2 mol of Al → ?? mol of Al 2 O 3  Moles of Al 2 O 3 = 2×2= 1 mol 4  Mw.t of Al 2 O 3 = 102 g/mol  Theoretical yield of Al 2 O 3 (grams of Al 2 O 3 )= n × M w.t  = 1× 102 = 102 g

15  B) Percent yield = Actual yield (g) x 100 Theoretical yield (g)  = 51 g × 100 = 50 % 102 g

16 Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C (g) + O 2 (g) → 2CO (g) What is the percent yield if 40.0 g CO are produced when 30.0 g of O 2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

17  Actual yield of CO = 40 g  Moles of O 2 = 30 g = 0.94 mol 32 g/mol  From the equation:  2C(g) + O 2 (g) → 2CO(g)  1 mol of O 2 → 2 mol of CO  0.94 mol of O 2 → ?? Mol of CO  Moles of CO = 0.94× 2 = mol 1

18  Theoretical yield of CO (m) = n × M w.t = × (12+16) = 52.5 g  Percent yield = 40 × 100 = 76.2% 52.5 So the correct answer is (3) 76.2%

19  Thank you


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