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Limiting/Excess Reagent Problem

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This resource details the stoichiometric calculations for a reaction between aluminum and lithium hydroxide. Given 12.3g of aluminum and 15.8g of lithium hydroxide, we determine moles and grams for each reactant, followed by identifying limiting and excess reagents. The balanced equation is provided, alongside calculations showing mass-to-mole conversions and verification of the law of conservation of mass. The limiting reagent in this reaction is aluminum, with lithium hydroxide as the excess reagent.

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Limiting/Excess Reagent Problem

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  1. Limiting/Excess Reagent Problem Stacy McFadden Pd. 7

  2. 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

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

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

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

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

  7. 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 1mole1 26.982g = .452 moles • 15.8 g15.8g x 1mole 1 23.9479 g =.660

  8. 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 g15.8g x 1mole 1 23.9479 g =.660

  9. 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 g15.8g x 1mole 1 23.9479 g =.660

  10. 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 1mole1 26.982g = .452 moles • 15.8 g15.8g x 1mole 1 23.9479 g =.660

  11. 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 g15.8g x 1mole 1 23.9479 g =.660

  12. 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 moles1 26.982g • 1.36moles x 23.9479g1.36moles 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 + 3.0237x 3 78.0027 • 3.0237 • 15.8 g15.8g x 1mole 1 23.9479 g =.660 moles

  13. 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 moles1 26.982g • 1.36moles x 23.9479g1.36moles 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 + 3.0237x 3 78.0027 3.0237 • 15.8 g44.8 g15.8g x 1mole 1 23.9479 g 44.74 g =.660 moles

  14. 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

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