Limiting Reagents Problem: Find the amount of cakes we can make when we have 4 sticks of butter and 50 cups of flour. 1 Butter + 2 Flour -> 1 Cake We need to find out which one of or reactants is our limiting reagent. So we do some calculations. If I have 4 sticks of butter, then I use up 8 cups of flour. Since I have 50 cups of flour, and I only used 8, then I have 42 cups of flour left over in excess reagent. Since flour is my excess, this means butter must be my limiting reagent. Now that I have found out that butter is my limiting reagent I will solve the problem with that knowledge. I have 4 sticks of butter, so I can only produce 4 cakes!

Limiting Reagents Determine the mass of P4O10 formed if 25g P4 and 50g O2 are combined. P4 + 5O2  P4O10 What is the given? We have to find the limiting reagent! 1 Mole P4 5 Mole O2 32 g O2 25 g P4 X X X = 32.26 g O2 124 g P4 1 Mole P4 1 Mole O2

Limiting Reagent 32.26 O2 was what we got when we used up all of our P4. We have to now compare how much we used, to how much we had. We had 50 g O2. We used 32.26 g O2. We therefore had 17.74 g O2 left over, in other words, in excess. Since we have O2 left over, but have used up all of our P4, we now know that O2 is our excess reagent. This means that P4 is our limiting reagent.

Limiting Reagent We now know to use P4, our limiting reagent as our given. This is because we can only produce as much product as our limiting reagent will allow. Think of it this way: We have 5 bicycle frames, and 20 wheels, how many bicycles can we produce? Frames: 5 bikes Wheels: 10 bikes We can only make 5 bikes, because frames are our limiting reagent!

Limiting Reagents Determine the mass of P4O10 formed if 25g P4 and 50g O2 are combined. P4 + 5O2  P4O10 The limiting reagent was P4, therefore we use P4 as our given to solve. 1 Mole P4 1 Mole P4O10 284 g P4O10 25 g P4 X X X = 57.26 g P4O10 124 g P4 1 Mole P4 1 Mole P4O10

Percent Yield Calculations If 100 g of Na and 100 g of Fe2O3 are used in this reaction, how much Na2O, can be made? 6Na + Fe2O3  Na2O + 2Fe 1 Mole Na 1 Mole Fe2O3 160 g Fe2O3 100 g Na X X X = 115.94 g Fe2O3 23 g Na 6 Mole Na 1 Mole Fe2O3 Fe2O3 is limiting

Percent Yield Calculations If 100 g of Fe2O3 and an excess of Na are used in this reaction, how much Na2O, can be made? 6Na + Fe2O3  Na2O + 2Fe 1 Mole Fe2O3 1 Mole Na2O 62 g Na2O 100 g Fe2O3 X X X = 38.75 g Na2O 160 g Fe2O3 1 Mole Fe2O3 1 Mole Na2O This is the theoretical yield!

Percent Yield Calculations Calculate the percent yield if the reaction actually yields 28.6 g of Na2O. 38.75 g Na2O was our theoretical. It is our theoretical because we had to do the math on paper. Actual 28.6 Theoretical 38.75 X 100 = X 100 = 73.81%

Using Volume With Stoichiometry What volume of NH3 at STP is produced if 30.0 g of N2 is reacted with an excess of H2? N2+3H2→ 2NH3

Using Volume With Stoichiometry How many liters of propane gas (C3H8) are needed to make 75g of water? C3H8+ 5O2 -> 3CO2+ 4H2O