Steps for solving Stoichiometric Problems Involving Solution Write the balanced equation for the reaction. Step 2 Calculate the moles of reactants Step 3 Calculate the limiting reactant Step 4 Calculate the moles of other reactants or products Step 5 Convert to grams or other units, if required.

Practice writing net ionic equations for these reactions. A net inoic equation only shows the components that are directly involved in the reaction. K2CrO4 (aq) + Ba(NO3)2 (aq)  BaCrO4 (s) + 2KNO3 (aq) CrO4- (aq) + Ba2+ (aq)  BaCrO4 (s) Na2SO4 (aq) + Pb(NO3)2 (aq)  PbSO4 (s) + 2NaNO3 (aq) SO42- (aq) + Pb2+ (aq)  PbSO4 (s) Na2S (aq) + NiCl2 (aq)  NiS (s) + 2NaCl (aq) S2- (aq) + Ni2+ (aq)  NiS (s) NaOH (aq) + FeCl3 (aq)  Fe(OH)3 (s) + 2NaCl (aq) 3OH- (aq) + Fe3+ (aq)  Fe(OH)3 (s)

Problem 1:. Calculate the mass of solid NaCl that must be added to 1 Problem 1: Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl. Calculate the mass of AgCl formed. Step 1: Write a balance equation Ag+(aq) + Cl-(aq)  AgCl(s) Step 2: Calculate the moles of reactants n = M  V = (0.100mol L-1)(1.50L) = 0.150 mol Ag+ Step 3: Determine which reactant is limiting We are adding just enough Cl- to react with the Ag+ present. So the Ag+ is the limiting reactant

Ag+(aq) + Cl-(aq)  AgCl(s) Step 4: Calculate the moles of Cl- required We have 0.150 mol of Ag+ ions and , because one Ag+ ion reacts with one Cl- ion, we need 0.150 mol of Cl- : Ag+(aq) + Cl-(aq)  AgCl(s) So 0.150 mol of AgCl will be formed. Step 5: Convert to grams of NaCl required To produce 0.150 mol Cl- , we need 0.150 mol of NaCl. We calculate the mass ofNaCl required as follows. Mass = n  Molar mass The mass of AgCl formed is = (0.150 mol)(58.4gmol-1) = 8.76 g NaCl Mass = n  Molar mass = (0.150 mol)(143.3gmol-1) = 21.5 g AgCl

Neutralization reactions An acid-base reaction is often called a neutralization reaction. This is because when you add just enough strong base to react exactly with a strong acid in a solution, we say the acid has been neutralized. One product of a neutralization reaction is always water. Acid + a base = water and a salt HCl (aq) + KOH (aq)  H20(l) + KCl (aq) The net reaction is: H+ (aq) + OH- (aq)  H20(l)

Problem 1:. Calculate the volume of a 0 Problem 1: Calculate the volume of a 0.100 M HCl solution needed to neutralize25.0 mL of a 0.350 M NaOH solution. Step 1: Write a balance equation H+(aq) + OH-(aq)  H2O(l) Step 2: Calculate the moles of reactants n = M  V = (0.350mol L-1)(0.025L) = 0.00875 mol OH- Step 3: Determine which reactant is limiting We are adding just enough H+ ions to react exactly with the OH- ions present. So the OH- is the limiting reactant

H+(aq) + OH-(aq)  H2O(l) Step 4: Calculate the moles of H+ required We have 0.00875 mol of OH- ions and , because one OH- ions reacts with one H+ ions, we need 0.00875 mol of H+: H+(aq) + OH-(aq)  H2O(l) So 0.00875 mol of H2O will be formed. Step 5: Calculate the volume of 0.100M HCl required To produce 0.150 mol Cl- , we need 0.150 mol of NaCl. We calculate the mass ofNaCl required as follows. Volume = n / molarity = (0.00875 mol)/(0.100molL-1) = 0.0875 L = 87.5 mL Therefore, 87.5mL of 0.100M HCl is required to neutralize 25.0 mL of 0.350 M NaOH