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Limiting Reagents

Limiting Reagents. If you are making tricycles, and you have 23 wheels , 20 handlebars , and 53 seats, how many tricycles can you make? 23 w (1 tricycle/ 3 w) = 7.7 tricycles 20 h (1 tricycle/ 2 h) = 10 tricycles 53 s (1 tricycle/ 1 s) = 53 tricycles

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Limiting Reagents

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  1. Limiting Reagents

  2. If you are making tricycles, and you have 23 wheels, 20 handlebars, and 53 seats, how many tricycles can you make? • 23 w (1 tricycle/ 3 w) = 7.7 tricycles • 20 h (1 tricycle/ 2 h) = 10 tricycles • 53 s (1 tricycle/ 1 s) = 53 tricycles • Which item limits the amount of tricycles that can be made? • What item(s) are in excess?

  3. Whatever substance is completely used up is always the limiting reagent. • Items that are left over are called excess reagents.

  4. Determining the Limiting Reagent 1. Make sure equation is balanced. 2. Find out how many moles of each reactant you have. 3. Find out how many moles are needed of each reactant using the mole ratio. 4. Identify limiting reagent. 5. You must begin every calculation using the amount of limiting reagent!

  5. 2 Cu + S  Cu2S • What is the limiting reagent when 80.0 g of Cu react with 25.0 g of S?

  6. What is the maximum number of grams of Cu2S that can be formed when 80.0 g of Cu reacts with 25.0 g of S?

  7. How many grams of the excess reagent does not react?

  8. Mg + 2 HCl  MgCl2 + H2 • Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg.

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