Limiting Reactant and Percent Yield Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how.

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Presentation transcript:

Limiting Reactant and Percent Yield

Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make. 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

Example u 2H 2 + O 2 → 2H 2 O u Suppose 2 moles of hydrogen pass through 2 moles of oxygen u 1 mole of oxygen will be extra and not used up in the reaction u The substance which is completely consumed is called the limiting reactant

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 u 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?

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

Your turn u If 10.1 g of magnesium and 4.67g of HCl gas are reacted, how many grams of gas will be produced? u How much excess reagent remains?

Your Turn II 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?

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

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?

Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %.

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