# Chapter 9 - Section 3 Suggested Reading: Pages

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Chapter 9 - Section 3 Suggested Reading: Pages 312 - 318
Stoichiometry Chapter 9 - Section 3 Suggested Reading: Pages

Limiting Reactants Available Ingredients 4 slices of bread
1 jar of peanut butter 1/2 jar of jelly Limiting Reactant bread Excess Reactants peanut butter and jelly

Limiting Reactants Available Ingredients Copper Wire 0.5 g AgNO3
0.5 grams AgNO3 Excess Reactants Copper Wire

Limiting Reactant The reactant that limits the amount of product that can be formed.

When quantities of reactants are available in the exact ratio described by the balanced equation, they are said to be in stoichiometric proportions.

Limiting Reactants Limiting Reactant used up in a reaction
determines the amount of all products formed Excess Reactant added to ensure that the other reactant is completely used up usually cheaper & easier to recycle

Solving Problems – Limiting Reactants
1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. The reactant that produces the smaller amount of product is the limiting reactant. Very similar to mass-mass problems!

Step 1: Write a balanced equation.
Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O H2  H2O

Step 2: For each reactant, calculate the amount of product formed. Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O H2  H2O

= O2 + 2H2  2 H2O wanted 2 X H2O = 1 given 0.038 mol X =
Step 2: 1.22 g oxygen 1 mole = 0.038 mol O2 32 g O H2  H2O given wanted wanted 2 X H2O = 1 given 0.038 mol X = 0.076 mol H2O

= O2 + 2H2  2 H2O wanted 2 X H2O = 2 given 0.525 mol X =
Step 2: 1.05 g H2 1 mole = 0.525 mol H2 2 g O H2  H2O given wanted wanted 2 X H2O = 2 given 0.525 mol X = 0.525 mol H2O

The one that produces the smallest amount is your limiting reactant.
Step 3: The one that produces the smallest amount is your limiting reactant. 1.22 g of O2 would produce mol H2O Oxygen is your limiting reactant! 1.05 g of H2 would produce mol H2O

Limiting Reactants Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

Step 1: Write a balanced equation.
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride. 2Na Cl2  2NaCl

Step 2: For each reactant, calculate the amount of product formed. Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride. 2Na Cl2  2NaCl

= 2Na + Cl2  2NaCl wanted 2 X NaCl = 2 given 0.0739 mol X =
Step 2: 1.7 g Na 1 mole = mol Na 23 g 2Na Cl2  2NaCl wanted given wanted 2 X NaCl = 2 given mol X = mol NaCl

= 2Na + Cl2  2NaCl wanted 2 X NaCl = 1 given 0.116 mol X =
Step 2: 2.6 L Cl2 1 mole = 0.116 mol Cl2 22.4 L 2Na Cl2  2NaCl wanted given wanted 2 X NaCl = 1 given 0.116 mol X = 0.232 mol NaCl

The one that produces the smallest amount is your limiting reactant.
*Step 3: The one that produces the smallest amount is your limiting reactant. 1.7 g Na would produce mol NaCl Sodium is your limiting reactant! 2.6 L Cl2 would produce mol NaCl

Percent Yield measured in lab calculated on paper

Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g Theoretical Yield: 45.8 g K2CO3 1 mol K2CO3 138 g 2 mol KCl 1 mol K2CO3 74 g KCl 1 mol KCl = 49.1 g KCl

Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.1 g
actual: 46.3 g Theoretical Yield = 49.1 g KCl 46.3 g 49.1 g % Yield =  100 = 94.3%

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