Limiting Reactants and Percent Yield

Transcription

1 Limiting Reactants and Percent Yield

2 What is a Limiting Reactant? o It is the reactant in a reaction that determines how much product can be made. o It is whatever reactant you have the least amount of. o If you are making a bicycle and you have all the parts to make 100 bikes, but only 4 wheels available, how many bikes can you make? o What is the limiting part? o For chemistry, it is whatever has the least amount of moles.

3 Limiting Reactant Rules To determine the limiting reactant: 1. Write/Confirm a balanced chemical equation. 2. Do a separate mass to mass problem for each reactant. The limiting reactant produces the least amount of moles or grams. To find out how much of the excess reactant is left over: 1. Start with the initial mass of the limiting reactant 2. Perform a mass to mass calculation to determine how much of the excess reactant was needed. 3. Subtract the excess value from the initial mass of the excess reactant.

4 Take the reaction: 4 NH O 2 4 NO + 6 H 2 O. In an experiment, 3.25 g of NH 3 are allowed to react with 3.50 g of O 2. Determining the Limiting Reactant Setup and workout Mass-Mass Problems for EACH reactant. Note: choose ONLY 1 product to convert both of your reactants into. NH g NH3 1 mole NH g NH3 4 mole NO 4 mole NH g NO 1 mole NO = 5.73 grams of NO produced O g O2 1 mole O g O2 4 mole NO 5 mole O g NO 1 mole NO = 2.63 grams of NO produced The limiting reactant is the reactant that produces the smallest amount of grams. The excess reactant is the reactant that produces the largest amount of grams.

5 Take the reaction: 4 NH O 2 4 NO + 6 H 2 O. In an experiment, 3.25 g of NH 3 are allowed to react with 3.50 g of O 2. Determining the Excess Reactant 1. Start with the initial mass of the limiting reactant O 2 is the limiting reactant. The initial mass of O 2 is 3.50 grams 2. Do a mass to mass problem to determine how much of the excess reactant was needed grams O 2 1 mole O 2 4 mole NH grams NH 3 = 1.49 grams of NH 3 used in reaction 31.99g O 2 5 mole O 2 1 mole NH 3 3. Subtract that value from the initial mass of the excess reactant. Formula: Initial Amount of Excess Amount of Excess Used = Excess Amount Remaining 3.25 g NH g NH 3 = 1.76 g of NH 3 remaining

6 Class Activity 1. Consider the following reaction: 2 Al + 6 HBr 2 AlBr H 2 a. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H 2 are formed? b. For the reactant in excess, how many grams are left over at the end of the reaction? 3. Consider the following reaction: 2 CuCl KI 2 CuI + 4 KCl + I 2 a. When 0.56 moles of CuCl 2 reacts with 0.64 moles of KI, how many moles of I 2 are formed? b. For the reactant in excess, how many moles are left over at the end of the reaction? 2. Consider the following reaction: 3 Si + 2 N 2 Si 3 N 4 a. When moles of Si reacts with moles of N 2, how many moles of Si 3 N 4 are formed? b. For the reactant in excess, how many grams are left over at the end of the reaction? 4. Consider the following reaction: 4 FeS O 2 2 Fe 2 O SO 2 a. When moles of FeS 2 reacts with 5.44 moles of O 2, how many moles of SO 2 are formed? b. For the reactant in excess, how many moles are left over at the end of the reaction?

7 Percent Yield Percent Yield describes how much product was actually made in the lab versus the amount that theoretically could be made. Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made. Theoretical Yield The maximum amount of product that could be formed from given amounts of reactants. (You get this from doing a mass to mass stoichiometry calculation!) Actual Yield The amount of product actually formed or recovered when the reaction is carried out in the laboratory. Actual Yield Theoretical Yield = Percent Yield

8 What is the % yield of H 2 O if 138 grams H 2 O is produced from 16 grams H 2 and excess O 2? 2 H 2 + O 2 2 H 2 O Determine actual and theoretical yield. Actual is given, theoretical is calculated Actual Yield 138 grams H 2 O Theoretical Yield This is the amount that you could produce theoretically 16 g H 2 1 mole H 2 2 mol H 2 O g H 2 O = 143 g H 2 O 2.02 g H 2 2 mol H 2 1 mol H 2

9 What is the % yield of H 2 O if 138 grams H 2 O is produced from 16 grams H 2 and excess O 2? Percent Yield Calculation Actual Yield Theoretical Yield x 100 = Percent Yield Actual Yield = 138 grams Theoretical Yield = 143 grams 138 g 143 g x 100 = 97%

10 Class Activity 2 FePO Na 2 SO 4 Fe 2 (SO 4 ) Na 3 PO 4 1. If I perform this reaction with 25 grams of iron (III) phosphate and an excess of sodium sulfate, how many grams of iron (III) sulfate can I make? 2. If 18.5 grams of iron (III) sulfate are actually made when I do this reaction, what is my percent yield? 3. If I do this reaction with 15 grams of sodium sulfate and get a 65.0% yield, how many grams of sodium phosphate will I make?