Chapter 11 Stoichiometry The study of quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.

Mole – Mole Relationship Use a balanced chemical equation to determine the mole – mole relationship of any substance. Mole – Mole relationship can be applied to any substance in the balanced equation. N2H4 + 2 H2O2  N2 + 4 H2O According to this equation:

Practice Mole-Mole Determine the number mole-mole relationship for Ag in the following equation: 3 AgNO3 + Al  Al(NO3)3 + 3 Ag

Mole – Mole Problems 1 mol Zn = 2 mol HCl How many moles of HCl are needed to react with 2.3 moles of Zn? 2 HCl + Zn  ZnCl2 + H2 1 mol Zn = 2 mol HCl

Practice Problems 1) How many moles of Na are needed to react with 55.5 moles of Cl2, in the production of salt? Na + Cl2  NaCl 2) How many moles of Al(OH)3 are produced when 2.53 moles of H2O react. Al2O3 + H2O  Al(OH)3

11-2 Solving Stoichiometry Problems Using all other conversions from chapter 10, along with the mole-mole to determine unknown amounts of each substance. 1 mol = 6.02 x 1023 particles 1 mol = 22.4L at STP 1 mol = formula mass(g)

Example: Mass - Mass Determine the mass of sodium hydroxide produced when .25g of sodium reacts with water. 2 Na + 2H2O  2NaOH + H2 .25g x g 2mol Na = 2 mol NaOH

Practice Problem #1 What mass of bromine, is produced when fluorine reacts with 3.55 moles of potassium bromide? F2 + 2 KBr  2 KF + Br2 3.55 mol x g

Mass – Volume : Example Find the mass of aluminum required to produce 1.32 L of hydrogen gas at STP. 2 Al + 3 H2SO4  Al2(SO4)3 + 3 H2 x g 1.32 L

Volume – Volume: Relationship The volume relationship of 2 gases at STP are directly proportional to the mole-mole relationship of those 2 gases. Why? All gases at STP occupy the same volume, 22.4L / mol.

Example: Volume - Volume What volume of hydrogen gas is necessary to produce 16.0L of ammonia (NH3)? N2 + 3 H2  2 NH3 x L 16.0 L

Chapter 11–3: Limiting and Excess Reactants Limiting reactant The reactant that limits the amount of product obtained. Completely consumed in the reaction. No trace remains in the product. Excess reactant The reactant that isn’t completely used in the reaction and remains after the reaction is over.

Example Problem: 1st determine the limiting reactant. Determine the mass of NaCl produced when 1.2 mol of Na react with .85 mol Cl2. 2 Na + Cl2  2 NaCl 1.2mol .85mol ?mol 1st determine the limiting reactant. Convert moles of one given to moles of another.

Determining Limiting and Excess Compare: Have vs. Need Have = Given amount Need = Calculated amount If Have > Need, excess reactant If Need > Have, limiting reactant

Comparing You have .85mol Cl2 you need .6mol Cl2. Determine the mass of NaCl produced when 1.2 mol of Na react with .85 mol Cl2. 2 Na + Cl2  2 NaCl 1.2mol .85mol ?g You have .85mol Cl2 you need .6mol Cl2. Chlorine is the excess, which means Sodium is the limiting.

Finishing the Problem Determine the mass of NaCl produced when 1.2 mol of Na react with .85 mol Cl2. 2 Na + Cl2  2 NaCl 1.2mol .85mol ?g You must begin with the limiting given to determine the unknown amount.

Practice Problem 3ZnO + 2Al(NO3)3  Al2O3 + 3Zn(NO3)2 Considering 25g of ZnO react with .555moles of Al(NO3)3. How many grams of Al2O3 will be produced? Determine limiting/excess reactant. Determine excess that will remain. Determine the unknown.

Percent Yield Yield = Product Actual Yield Amount of product given in the problem. Theoretical Yield (Expected) Amount of product determined by a stoichiometry calculation. Percent Yield

Sample Problem Determine the percent yield when 2.80g Al(NO3)3 react to produce .966g of Al(OH)3. Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3 2.80g .966g

Identifying what to do! 1) Regular stoichiometry: 2) Limiting/Excess: Given one value. 2) Limiting/Excess: Given 2 values both are reactants (left side of equation) 3) %yield: Given 2 values 1 reactant and 1 product All problems involve a mole-mole conversion.

Practice Problem Determine the percent yield when 1.50g Al(OH)3 decomposes to produce .25g of H2O. 2Al(OH)3  Al2O3 + 3H2O 1.50g .25g