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Solutions Stoichiometry

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Solutions Stoichiometry Powered By Docstoc
					Steps for solving Stoichiometric Problems Involving Solution

Step 1 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)
S2- (aq) + Ni2+ (aq)  NiS (s)

Na2S (aq) + NiCl2 (aq)  NiS (s) + 2NaCl (aq)

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 1.50 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. Write a balance equation Step 1:

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

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 (0.150 mol)(58.4gmol-1) 8.76 g NaCl The mass of AgCl formed is
-1 Mass = n  Molar mass (0.150 mol)(143.3gmol= )21.5 g AgCl =

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 0.100 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 = 0.00875 mol OHn=MV = (0.350mol L-1)(0.025L)

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

Step 4: Calculate the moles of H+ required We have 0.00875 mol of OH- ions and , because one OHions 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. = 0.0875 L Volume = n / molarity= (0.00875 mol)/(0.100molL-1) = 87.5 mL Therefore, 87.5mL of 0.100M HCl is required to neutralize 25.0 mL of 0.350 M NaOH


				
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