Stoichiometry – Ch. 12.3
What would be produced if two pieces of bread and a slice of salami reacted together? + ?
2B + S
+
+ 2B + S B 2 S
+ 2B + 1S 1B 2 S
+ Bread (B) and Salami (S) react in a 2 to 1 (2:1) ratio to produce sandwich (B2S)
+ 6B + 2S ?
+
+ + (leftover)
6B + 2S + 2B 2 S
6B + 2S + 2B 2 S The amount of salami limits the amount of sandwiches that can be produced
WHY?
Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly
Limiting Reactant used up in a reaction determines the amount of product Excess Reactant Left over in a reaction added to ensure that the other reactant is completely used up
2H O 2 2H 2 O
? what happens if there is an excess amount of hydrogen?
?
?
? 5H O 2 2H 2 O+ 3H 2
1. Write a balanced equation. 2. For each reactant, calculate mol present and find ratio. 3. Smaller answer indicates: limiting reactant amount of product 4. Use limiting reactant to find mol of product and needed reactant 5. Convert mol to grams
Determining limiting reactant Calculating product Determining excess reactant
Determine the mass of tetraphosphorous decoxide formed if 25.0 g of phosphorus (P 4 ) and 50.0 g of oxygen are combined? First, write the equation… P 4 + O 2 P 4 O 10 Then balance it… P 4 + 5O 2 P 4 O 10
Determine moles of each reactant 25.0 g P 4 x 1 mol = mol P g 50.0 g O 2 x 1 mol = 1.56 mol O g 25.0 g50.0 g
Calculate mole ratio 1.56 mol O 2 = 7.72 mol O mol P 4 1 mol P 4 Determine mol ratio from equation 5 mol O 2 1 mol P 4
Because 7.72 mol O 2 is available but only 5 mol is needed to react with 1 mol P 4, O 2 is in excess and P 4 is the limiting reactant. Use mol of P 4 to determine mol of P 4 O 10 produced. Answer #1: P is limiting and O is excess Multiply limiting reactant (P 4 ) by mole ratio mol P 4 x 1 mol P 4 O 10 = mol P 4 O 10 1 mol P 4
Convert mol P 4 O 10 to grams mol P 4 O 10 x g P 4 O 10 = 57.3 g P 4 O 10 1 mol P 4 O 10 Answer #2: 57.3 g P 4 O 10 is produced
O 2 is in excess so only so much is used. Use P 4 to determine mol and mass of O 4 used mol P 4 x 5 mol O 2 = 1.01 mol O 2 needed 1 mol P 4 Multiply mol O 2 by molar mass 1.01 mol O 2 x 32.0 g O 2 = 32.3 g O 2 needed 1 mol O 2
Subtract the mass of O 2 needed from the mass available to calculate excess O g O 2 available – 32.3 g needed = 17.7 g O 2 Answer #3: 17.7 g O 2 is left over (in excess)
6Na + Fe 2 O 3 3Na Fe Find: The limiting reactant The reactant in excess The mass of solid iron produced The mass of excess reactant that remains
calculated on paper measured in lab
When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl 2KCl + H 2 O + CO g? g actual: 46.3 g
Theoretical Yield: 45.8 g K 2 CO 3 1 mol K 2 CO g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO g KCl 1 mol KCl K 2 CO 3 + 2HCl 2KCl + H 2 O + CO g? g actual: 46.3 g
Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g 100 = 93.7% K 2 CO 3 + 2HCl 2KCl + H 2 O + CO g49.4 g actual: 46.3 g