Limiting/Excess Reagent Problem

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Presentation transcript:

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 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 1mole 1 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 1mole 1 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 moles 1 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 - 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?? 32.6 16.8 g - 15.8 16.8g