Limiting Reactant and Reactions in Aqueous conditions

• Stoichiometric Mixture

  Contains the relative amounts of reactants that matches the numbers in the balanced equation.

  N2 (g) + 3H2 (g) → 2NH3 (g)
• Limiting Reactant Mixture

  N2 (g) + 3H2 (g) → 2NH3 (g)

  The limiting reactant in a chemical reaction is the reactant that is completely consumed when the reaction is completed. In other words, it’s the reactant that determines the amount of product that can be formed because the reaction cannot continue without it.
• Calculations involving a limiting reactant

  • Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.
  • Let’s look at an example: Methane and water will react to form products according to the equation: CH4 + H2O → 3H2 + CO
  • The amount of products that can form is limited by the water.
  • Water is the limiting reactant.
  • Methane is in excess.
  • H2O molecules are used up first, leaving two CH4 molecules unreacted.
• Steps for Solving Stoichiometry Problems Involving Limiting Reactants

  1. Write and balance the equation for the reaction.
  2. Convert known masses of reactants to moles.
  3. Using the appropriate mole ratios, compute the numbers of moles of product formed if each reactant were consumed
  4. Choose the least number of moles of product formed from Step 3.
  5. Convert from moles of product to grams of product, using the molar mass (if this is required by the problem).
• Example of Solving a Stoichiometry Problem: Step 1

  You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B → 2C

  Where are we going?

  ▪ To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B.

  What do we need to know?

  ▪ The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation.

  ▪ The molar masses of A, B, and the product they form.
• Example of Solving a Stoichiometry Problem: Step 2


  You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B → 2C

  ▪ Convert known masses of reactants to moles.

  Moles A = (mass A) / (molar mass A)

  = 10 /10

  = 1 mole

  Moles B = (mass B) / (molar mass B)

  = 10 /20

  = 0.5 mole

  Work out the number of moles of product formed from each reactant.

  
• Example of Solving a Stoichiometry Problem: Step 3


  You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B → 2C

  ▪ Convert from moles of C to grams of C using the molar mass.

  mass C = (moles of C) x (molar mass C)

  = 0.333 moles C x 25.0 g/mol

  = 8.33 g of product C
• Reactions in solutions

  Macroscopic observation – indications of a chemical reaction could include:

  • a change in colour

  • a change in temperature

  • the evolution of heat

  • a change of phase (e.g. precipitation of a solid out of a solution)

  • the emission of light.
• Reactions in which a solid forms

  A reaction in which a solid forms is called a precipitation reaction. – Solid = precipitate

  • K2 CrO4 (aq) + Ba(NO3 ) 2 (aq) → Products

  
• Reactions in which a solid forms 2

  • K2 CrO4 (aq) + Ba(NO3 ) 2 (aq) → Products • The mixed solution contains four types of ions: K+ , CrO4 2– , Ba2+, and NO3 – .
  • Determine the possible products from the ions in the reactants. The possible ion combinations are:

  
• How to Decide What Products Form

  Decide which is most likely to be the yellow solid formed in the reaction.

  • K2CrO4 (aq) reactant
  • Ba(NO3 )2 (aq) reactant
  • The possible combinations are KNO3 and BaCrO4 .

  ▪ KNO3 white solid

  ▪ BaCrO4 yellow solid
• Using solubility rules


  .

  
• Predicting precipitates

  1. Write the reactants as they actually exist before any reaction occurs. Remember that when a salt dissolves, its ions separate.
  2. Consider the various solids that could form. To do this, simply exchange the anions of the added salts.
  3. Use the solubility rules to decide whether a solid forms and, if so, to predict the identity of the solid
• Spectator Ions and Net Equations

  • Spectator ions are aqueous ions that don’t participate in reactions: 2AgNO3 (aq) + CaBr2 (aq) → 2AgBr(s) + Ca(NO3 )2 (aq)
  • The NO3 – and Ca2+ ions don’t take part in the reaction.
  • What is the net ionic equation? 2Ag+ (aq) + 2Br- (aq) → 2AgBr(s)
• Summary

  • Limiting reactant (reagent) is the reactant that runs out first and thus limits the amounts of product(s) that can form.
  • There are different strategies that can be used to solve problems involving a limiting reagent
  • The importance of this is that not all of the reactant will be used up. Some will be in excess.
  • During reactions in aqueous solution, reagents that take part in chemical changes will be included in the net ionic reaction