LIMITING REAGENT. Taking Stoichiometric conversions one step further

Transcription

1 LIMITING REAGENT Taking Stoichiometric conversions one step further

2 Limiting Reagent The reactant that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed The only way to know which reactant is limiting is to compare the number of moles of each reactant Whichever is smaller is the limiting reactant

3 Steps to Calculating Limiting Reagent 1. Convert each reactant listed into moles of the same product The smallest is the limiting reactant! 2. Use the limiting reactant to calculate the yield of the product indicated in the questions

4 Example #1 Using the equation: 2Na + Cl 2 à 2NaCl Suppose that 6.70g of Na reacts with 3.20g of Cl 2. Which is the limiting reactant? How many grams of NaCl will be produced?

5 Example #1 Step 1: Calculate moles of reactant for each reactant 6.70 g Na X 1 mol Na = mol Na g Na 3.20 g Cl 2 X 1 mol Cl2 = mol Cl g Cl mol of Cl 2 is smaller than mol of Na Therefore, Cl 2 is the limiting reagent

6 Example #1 Using the equation: 2Na + Cl 2 à 2NaCl Suppose that 6.70g of Na reacts with 3.20g of Cl 2. Which is the limiting reactant? How many grams of NaCl will be produced?.045 mol Cl 2 2 mol NaCl g NaCl X X = 1 mol Cl 2 1 mol NaCl 5.26 g NaCl

7 Example #2 Suppose that 80.0g of copper reacted with 25.0g of sulfur. What is the limiting reactant? What is the maximum amount of copper (I) sulfide that can be produced? This problem is slightly different in that you need to write the balanced equation! 2Cu + S à Cu 2 S

8 Example #2 Step 1: Calculate moles of product from each reactant 80.0 g Cu X 1 mol Cu = 1.26 mol Cu g Cu 25.0 g S X 1 mol Cu = 0.78 mol S g S Limiting Reagent: Sulfur

9 Example #2 Suppose that 80.0g of copper reacted with 25.0g of sulfur. What is the limiting reactant? What is the maximum amount of copper (I) sulfide that can be produced?.78 mol S 1 mol Cu 2 S g Cu 2 S X X = 1 mol S 1 mol Cu 2 S g Cu 2 S

10 Leftovers Sometimes you will be asked to determine how much of the excess reactant you ll have left For excess, you will need an additional 2 steps 1. Calculate the amount of the actual reactant needed using the theoretical yield calculated from the limiting reactant 2. Subtract the difference of what was needed in the excess reactant and the amount actually in the problem

11 Example #3 Suppose 25.5g of Calcium carbide, CaC 2, reacts with 15.5g of water. What is the maximum amount of acetylene, C 2 H 2, that can be produced? How much of the excess reactant will you have left over? CaC 2 + 2H 2 O à C 2 H 2 + Ca(OH) 2

12 Example #3 Step 1: Determine the limiting reagent 25.5 g CaC 2 X 1 mol CaC 2 = mol CaC g CaC g H 2 O X 1 mol H 2 O = 0.86 mol H 2 O g H 2 O Limiting reagent is CaC 2 and Excess reagent is H 2 O

13 Example #3 Suppose 25.5g of Calcium carbide, CaC 2, reacts with 15.5g of water. What is the maximum amount of acetylene, C 2 H 2, that can be produced? How much of the excess reactant will you have left over?

14 Example #3 Step 2: Convert moles of limiting reagent to grams of leftover reagent.398 mol CaC 2 2 mol H 2 O g H 2 O X X = 1 mol CaC 2 1 mol H 2 O g H 2 O

15 Example #3 This means that we will only make g of H 2 O. We had 15.5 g. So how much do we have left? 15.5 g = g 1.16 g H 2 O Left Starting Amount Amount Needed

16 Percent Yield So far, every problem we have calculated with stoichiometry has used a theoretical value This is what we SHOULD get under ideal conditions However, in the lab we don t usually get that amount L In order to know how successful the experiment was, we need to account for the amount that we got versus what we should have gotten

17 Percent Yield Percent yield: The ratio of the actual yield to the theoretical yield Expressed as a percentage % Yield = Actual Yield X 100 Theoretical Yield

18 Percent Yield Problems Must tell you two out of 3 pieces of information: The amount of at least one reactant The amount of product that was produced The percent yield You will need to use the amount of the reactant given to calculate the theoretical yield, then plug into the % yield equation

19 Percent Yield Example #1 When 29.5g of CaCO3 is heated it decomposes into CaO and CO g of CaO is actually produced. What is the percent yield of CaO in this reaction? CaCO 3 à CaO + CO g CaCO 3 x 1 mol CaCO 3 x 1 mol CaO x g CaO = g CaCO 3 1 mol CaCO 3 1 mol CaO 14.2g CaO x 100% = 85.90% 16.53g CaO 16.53g CaO