Limiting/Excess Reagent Problem

Stacy McFadden Pd. 7

Calculate the moles and masses(g) for each chemical in the reaction, if 12.3 g of Aluminum is reacted with 15.8 g

of Lithium Hydroxide

Write a complete and balanced equation

Al + 3Li(OH) 3Li + Al(OH)3

Draw a column for each chemical

Al + 3Li(OH) 3Li + Al(OH)3

Draw a line to separate the columns into two rows

Al + 3Li(OH) 3Li + Al(OH)3

Write the amounts given in the appropriate columns

Al + 3Li(OH) 3Li + Al(OH)3

12.2 g

15.8 g

Convert the amounts given into moles. Since Al has less moles than Li(OH), You only have to worry about the top half

Al + 3Li(OH) 3Li + Al(OH)3

12.2 g 12.2g x 1mole

1 26.982g

= .452 moles

15.8 g 15.8g x 1mole 1 23.9479 g

=.660

In each of the other columns write the moles of given (x) a fraction

Al + 3Li(OH) 3Li + Al(OH)3

12.2 g .452 moles x / .452 moles x / .452 moles x /

12.2g x 1mole1 26.982g

= .452 moles

15.8 g 15.8g x 1mole 1 23.9479 g =.660

The Numerator of the fraction is the coefficient of that column

Al + 3Li(OH) 3Li + Al(OH)3

12.2 g .452 moles x 3/ .452 moles x 3/ 452 moles x 3/

12.2g x 1mole1 26.982g

= .452 moles

15.8 g 15.8g x 1mole 1 23.9479 g

=.660

The denominator of the fraction is the coefficient of the given column

Al + 3Li(OH) 3Li + Al(OH)3

12.2 g .452 moles x 3/1 .452 moles x 3/ 1 .452 moles x 1/1 12.2g x 1mole

1 26.982g

= .452 moles

15.8 g 15.8g x 1mole 1 23.9479 g

=.660

Do math and label as moles

Al + 3Li(OH) 3Li + Al(OH)3

12.2 g .452 moles x 3/1 .452 moles x 3/ 1 .452 moles x 1/1

12.2g x 1mole = 1.36 moles = 1.36 moles = .452 moles1 26.982g

= .452 moles

15.8 g 15.8g x 1mole 1 23.9479 g

=.660

Covert all moles to grams

Al + 3Li(OH) 3Li + Al(OH)3 12.2 g .452 moles x 3/1 .452 moles x 3/1 .452 moles x 1/1 12.2g x 1mole = 1.36 moles = 1.36 moles = .452 moles

1 26.982g 1.36moles x 23.9479g 1.36moles x 6.941g .452 moles x 78.0027g

1 1mole 1 1mole 1 1mole = .452 moles 6.941 = 32.6 g = 9.44 g 15.999 = 35.3 g + 15.999 x 3 1.0079 47.997 26.982 23.9479 47.997

1.0079 + 3.0237 x 3 78.0027

3.0237

15.8 g 15.8g x 1mole 1 23.9479 g =.660 moles

Verify the law of conservation of mass

Al + 3Li(OH) 3Li + Al(OH)3 12.2 g .452 moles x 3/1 .452 moles x 3/1 .452 moles x 1/1 12.2g x 1mole = 1.36 moles = 1.36 moles = .452 moles

1 26.982g 1.36moles x 23.9479g 1.36moles x 6.941g .452 moles x 78.0027g

1 1mole 1 1mole 1 1mole = .452 moles 6.941 = 32.6 g = 9.44 g 15.999 = 35.3 g + 15.999 x 3 1.0079 47.997 26.982 23.9479 47.997

1.0079 + 3.0237 x 3 78.0027 3.0237

15.8 g 44.8 g 15.8g x 1mole 1 23.9479 g 44.74 g =.660 moles

ANSWER• 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?? 32.6 16.8 g

- 15.8 16.8g