Chapter 7: Chemical Reactions and Stoichiometry

  • Definition of Chemical Reactions:

 - Chemical and physical changes that matter undergoes are represented by chemical equations using the formulas of the reactants and the products, whether they are elements or compounds.

  • Classification of Reactions:

 - Chemical reactions can be classified in various ways. One classification method is as follows:

 - Combination Reaction:

 - Example: Ca(s) + Cl2(g)

 - Combustion Reaction:

 - Example: 4Fe(s) + 3O2(g)
 
  - A combustion reaction involves oxygen as one of the reactants.

 - Dissolution (Physical Change):

 - Example: C6H12O6(s) + H2O
 

 - Decomposition Reaction:

 - Example: CaCO3(s) + heat
 

 - Distinction Between Chemical Change and Physical Change:

 - A chemical change results in the formation of new substances, while a physical change does not.

7-2 Balancing Chemical Equations

  • Conservation of Mass:

 - The total mass of reactants equals the total mass of the products.

  • Balancing Chemical Equations:

 - The equation 2 Na + Cl2
 

 - Example of combustion of C3H6:

 - The unbalanced reaction: C3H6 + O2
 
 

 - Balancing Process:

 1. Balance Carbon:

 - C3H6 + O2
 

 2. Balance Hydrogen:

 - C3H6 + O2
 
 
 

 3. Balance Oxygen:

 - 9 oxygen in the product leads to C3H6 + 4.5 O2
 
 
 

 4. Change fraction to an integer:

 - Multiply all coefficients by 2, yielding 2 C3H6 + 9 O2
 
 
 

 - Reaction Media:

 
 - Reactions can occur in various media:

 1. Solution

 2. Gas Phase

 3. Solid Phase

 4. Combination of different phases

 - Aqueous Solution:

 - An aqueous solution contains a solute (the smaller amount) dissolved in a solvent (the larger amount), which is water.

7-3 Reaction Stoichiometry

  • Stoichiometry Definition:

 - Stoichiometry requires a balanced equation that provides the molar relationship between the products and the reactants and their quantities.

  • Information from a Balanced Equation:

 - Example: 2 H2(g) + 1 O2(g)
 
 
 
 

 - Molar Masses:

 - 2 mol x (2.02 g/mol) = 4.04 g (for H2)

 - 1 mol x (32.0 g/mol) = 32.0 g (for O2)

 - 2 mol x (18.02 g/mol) = 36.04 g (for H2O)

  • Stoichiometric Calculations Example:

 - Given: 2 H2(g) + O2(g)
 
 
 

 - Question: How many moles of oxygen are needed to completely react with 5.0 moles of hydrogen?

  • Calculation:

 - From the balanced equation, 1 mole of O2 is required for every 2 moles of H2.

 - moles O2 needed = 5.0 moles H2 x (1 mol O2 / 2 mol H2) = 2.5 moles of O2.

7-4 Further Stoichiometric Calculations

  • Question: Calculate the amount of H2O produced in the given example.

  • Answer:

 - 5 moles of H2 produce 5 moles of H2O.

  • Caloric Calculations:

  • Example Reaction Calculation:

 - Fe2O3 + 3 CO
 
 
 

 - a. How many moles of CO are needed to completely react with 40.0 g of Fe2O3?

 - b. How many grams of CO2 are produced in the reaction?

  • Solution for (a):

 - Convert the mass of reactant Fe2O3 to moles:

 - Molecular weight = 160 g/mol

 - 40.0 g Fe2O3 x (1 mol / 160 g) = 0.250 moles of Fe2O3.

 - Number of moles of CO needed:

 - 0.250 moles Fe2O3 x (3 moles CO / 1 mole Fe2O3) = 0.750 mol CO.

  • Solution for (b):

  • Step 1: Number of moles of CO2 produced:

     - 0.250 moles Fe2O3 x (3 moles CO2 / 1 mole Fe2O3) = 0.750 moles CO2.

  • Step 2: Convert the number of moles of CO2 to mass:

     - 0.750 moles CO2 x 44.0 g/mol = 33.0 g CO2.

  • Summary:

 - mass of reactant (Fe2O3) giving moles, and corresponding product mass:

 - 2 moles H2O
 
 
 

 - 18.02 g H2O
 

 - 1 mol
 

 - 1 mol Fe2O3
 

 - 160 g
 

 - 3 moles CO
 

 - 1 mole Fe2O3
 

 - 3 moles CO2
 

 - 1 mole Fe2O3
 

 - 44.0 g CO2
 

 - 1 mole CO2

7-5 Limiting Reagent (L.R.)

  • Definition of Limiting Reagent:

 - The limiting reagent is the reactant that is consumed entirely during the reaction, thus governing the quantities of the products obtained.

  • Reactions with Stoichiometric Amounts:

 - If reactants are mixed in stoichiometric amounts, they are completely consumed, with no excess reactants left over.

  • Determining the Limiting Reagent:

 - Example: Assembling a bicycle requires 2 wheels (W) and 1 handlebar (H):

 - 2 W + 1 H
 
 
 

 - If starting with 8 wheels and 10 handlebars:

 - If all 8 wheels are used, 4 handlebars are needed. Since 10 handlebars are available, those not consumed represent an excess.

 - Thus, the limiting reagent is W.

  • Another Example:

 - For 2 Na + Cl2
 
 
 

 - If all Sodium (Na) reacts:

 - 5.0 moles Na
 
 
 

 - 5.0 moles Cl2 are available.

 - Therefore, Cl2 is in excess (2.5 moles) and Na is the limiting reagent.

7-6 Methods to Find the Limiting Reagent

  • Method 1:

 - For a reaction A + B
 
 
 

 - Use the amount of reactant A to calculate the amount of B needed using stoichiometry.

 - a) If more B is available than needed, then B is in excess, and A is limiting.

 - b) If the amount of B calculated is more than what is supplied, then B is the limiting reagent.

  • Example: For the reaction P4 + 6 Cl2
     
     
     

 - If all P4 reacts:

 - 6 moles of Cl2 are needed for every mole of P4, hence:

 - 5 moles P4 x 6 = 30 moles Cl2 needed.

 - However, only 10 moles of Cl2 are provided, so Cl2 is the limiting reagent in this case.

7-7 Another Method for Finding Limiting Reagent

  • Method 2:

 - Identify the limiting reagent by calculating the product yielded from each reactant.

 - The reactant that yields the least amount of product is the limiting reagent.

  • Example Calculation: For P4 + 6 Cl2
     
     
     

 - If all P4 reacts:

 - 5 moles P4 x (4 moles PCl3 / 1 mole P4) = 20 moles PCl3.

 - If all Cl2 reacts:

 - 10 moles Cl2 x (4 moles PCl3 / 6 moles Cl2) = 6.7 moles PCl3.

 - Comparing the results, the limiting reagent is Cl2, since it produces the least amount of PCl3 (6.7 moles).

  • Notes on Reagents:

 - Once identified, subsequent calculations should be based on the limiting reagent, including theoretical yields, amount of product, and the amounts of excess reactants left over.

 - The limiting reagent is fully consumed by the end of the reaction, resulting in zero moles remaining.

7-8 Reaction Yield

  • Percent Yield Definition:

 - Most reactions conducted in the lab do not go to completion; not all reactants yield products as indicated by the balanced equation.

 - The calculated amount of product from stoichiometric calculations is referred to as the theoretical yield, while the quantity obtained through experimental work is the actual yield.

  • Formula for Percent Yield:

 - Percent Yield = (Actual Yield / Theoretical Yield) x 100.

  • Example with Reaction of NH3 and CuO:

 - Balanced reaction: 2 NH3(g) + 3 CuO(s)
 
 
 

 - Given:

 - 18.1 g NH3

 - 90.4 g CuO

 - Isolate only 8.0 g of N2(g).

7-9 Calculating Yield Example Steps

  • Steps to Determine L.R., Theoretical Yield, and Percent Yield:

     1) Convert Mass to Moles:

  • For NH3:

     - 18.1 g NH3 x (1 mol / 17.0 g) = 1.06 mol NH3.

  • For CuO:

     - 90.4 g CuO x (1 mol / 79.5 g) = 1.14 mol CuO.

     2) Rewriting the Equation with Moles:

     - 2 NH3(g) + 3 CuO(s)
     
     
     

 - Starts with: 1.06 moles NH3 and 1.14 moles CuO.

 3) Calculate Product Yield:

 - If all CuO were to react:

 - 1.14 mol CuO
 
 
 

 - If all NH3 were to react:

 - 1.06 mol NH3
 
 
 

 - Therefore, CuO is the limiting reagent as it produces the least amount of N2 (0.380 mol).

 4) Finding Theoretical Mass of N2:

 - Use the amount of N2 yielded:

 - 0.380 moles N2 x 28.0 g/mol = 10.6 g N2.

 5) Finding Mass of Excess Reagent:

 - To find the mass of the excess NH3:

 - 1.14 mol CuO
 
 
 

 - Calculate excess:

 - 1.06 mole - 0.760 mole = 0.30 moles NH3 yielding mass:

 - 0.30 mol NH3 x 17.0 g/mol = 5.1 g NH3.

 6) Finding Percent Yield:

 - Percent yield calculation:

 - Percent Yield = (8.0 g N2 / 10.6 g N2) x 100 = 75.5.