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Limiting Reagent u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent.

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Presentation on theme: "Limiting Reagent u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent."— Presentation transcript:

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2 Limiting Reagent u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent determines how much product you can make

3 How do you find out? u Do two stoichiometry problems. u The one that makes the least product is the limiting reagent. u For example Copper reacts with sulfur to form copper(I) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

4 u If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?  2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

5 How much excess reactant? u Use the limiting reactant to find out how much excess reactant you used u Subtract that from the amount of excess you started with

6 Your turn  Mg(s) +2 HCl(g)  MgCl 2 (s) +H 2 (g) u If 4.87 mol of magnesium and 9.84 mol of HCl gas are reacted, how many moles of gas will be produced? u What is the limiting reagent? u How much excess reagent remains?

7 Your Turn u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? u How much excess reagent will remain?

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9 Yield u The amount of product made in a chemical reaction. u There are three types u Actual yield- what you get in the lab when the chemicals are mixed u Theoretical yield- what the balanced equation tells you you should make. u Percent yield u Percent yield = Actual x 100 % Theoretical

10 Example u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield? u If you had started with 9.73 g of Al, how much copper would you expect to produce?

11 Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %. u How would you get good at this?


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