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Limiting/Excess Reagent Problem

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Presentation on theme: "Limiting/Excess Reagent Problem"— Presentation transcript:

1 Limiting/Excess Reagent Problem
Stacy McFadden Pd. 7

2 Calculate the moles and masses(g) for each chemical in the reaction, if g of Aluminum is reacted with 15.8 g of Lithium Hydroxide

3 Write a complete and balanced equation
Al Li(OH)  3Li Al(OH)3

4 Draw a column for each chemical
Al Li(OH)  3Li Al(OH)3

5 Draw a line to separate the columns into two rows
Al Li(OH)  3Li Al(OH)3

6 Write the amounts given in the appropriate columns
Al Li(OH)  3Li Al(OH)3 12.2 g 15.8 g

7 Convert the amounts given into moles
Convert the amounts given into moles. Since Al has less moles than Li(OH), You only have to worry about the top half Al Li(OH)  3Li Al(OH)3 12.2 g 12.2g x 1mole g = .452 moles 15.8 g g x 1mole g =.660

8 In each of the other columns write the moles of given (x) a fraction
Al Li(OH)  3Li Al(OH)3 12.2 g moles x / moles x / moles x / 12.2g x 1mole g = .452 moles 15.8 g g x 1mole g =.660

9 The Numerator of the fraction is the coefficient of that column
Al Li(OH)  3Li Al(OH)3 12.2 g moles x 3/ moles x 3/ moles x 3/ 12.2g x 1mole g = .452 moles 15.8 g g x 1mole g =.660

10 The denominator of the fraction is the coefficient of the given column
Al Li(OH)  3Li Al(OH)3 12.2 g moles x 3/ moles x 3/ moles x 1/1 12.2g x 1mole g = .452 moles 15.8 g g x 1mole g =.660

11 Do math and label as moles
Al Li(OH)  3Li Al(OH)3 12.2 g moles x 3/ moles x 3/ moles x 1/1 12.2g x 1mole = moles = 1.36 moles = .452 moles g = .452 moles 15.8 g g x 1mole g =.660

12 Covert all moles to grams
Al Li(OH)  3Li Al(OH)3 12.2 g moles x 3/ moles x 3/ moles x 1/1 12.2g x 1mole = moles = 1.36 moles = .452 moles g 1.36moles x g moles x g moles x g mole mole mole = .452 moles = 32.6 g = g = g x x 3.0237 15.8 g g x 1mole g =.660 moles

13 Verify the law of conservation of mass
Al Li(OH)  3Li Al(OH)3 12.2 g moles x 3/ moles x 3/ moles x 1/1 12.2g x 1mole = moles = 1.36 moles = .452 moles g 1.36moles x g moles x g moles x g mole mole mole = .452 moles = 32.6 g = g = g x x 15.8 g g g x 1mole g g =.660 moles

14 ANSWER - Li(OH) Which Chemical is the Limiting Reagent?
(Given chemical in the smaller set of stoichiometry) Al Which Chemical is the Excess Reagent? (Given Chemical in the larger set of stoichiometry) - Li(OH) What is the Amount of Excess?? g g


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