Chemistry Notes: Reactions in Solutions
• Two solutions can be combined to generate a chemical reaction.
• Often, chemical reactions occur in aqueous* solutions, e.g. living systems.
• The concentration of each solution and the types of solutes involved must
be taken into account.
* Water = solvent
------------------------------------------------------------------------------------Example of a chemical reaction:
HCl + NaOH --> NaCl + H2O.
Let’s suppose this reaction occurs as a result of combining two solutions: HCl
& NaOH.
Let: HCl solution = 1.0 M (= 1.0 mol HCl/L)
NaOH solution = 1.0 M (= 1.0 mol NaOH/L)
Combine 10 mL of each solution together. How many grams of NaCl are
produced?
1st: Determine how many moles of each reactant are used.
10 mL of 1.0 M HCl = 10 mL x 1.0 mol = 0.01 mol HCl
1000
L
10 mL of 1.0 M HCl = 10 mL x 1.0 mol = 0.01 mol NaOH
1000
L
Divide mL by
1000 to
convert to L.
Next, determine moles of product produced (stoichiometry).
HCl + NaOH --> NaCl + H2O
0.01 +
mol
0.01
mol
--> 0.01
mol
0.01 mol of NaCl produced.
Then, convert moles to grams.
(0.01 mol)(23+35)g/mol = 0.58 g NaCl produced
Molar mass
NaCl
Example 2:
15.0 mL of a 0.25 M solution of AgNO3 is combined with 10.0 mL of a
0.35 M solution of KBr in the following reaction:
AgNO3 + KBr ---> AgBr + KNO3 (balanced)
1) What is the limiting reagent?
2) How many grams of AgBr are produced?
• Step 1: Determine the moles of each reactant used.
AgNO3 = (15 mL/1000)(0.25
KBr = (10 mL/1000)(0.35
mol/ )
L
mol/ )
L
= 0.00375 mol AgNO3
= 0.0035 mol KBr
• Step 2: Determine limiting reagent.
AgNO3 + KBr ---> AgBr + KNO3
Molar ratios are all 1, so the limiting reagent will be determined by the
smaller quantity of reactant, which is KBr @ 0.0035 mol.
• Step 3: Determine the moles of product.
0.0035 mol KBr ---> 0.0035 mol AgBr
• Step 4: Calculate the grams of AgBr.
(0.0035 mol AgBr)(108+80)g/mol = 0.66 g AgBr produced
You do one.
50 mL of a 5.00 M solution of Li2SO4 is combined with 45 mL of a 4.50 M
solution of MgCl2, according to the following reaction:
Li2SO4 + MgCl2 ---> 2LiCl + MgSO4.
 How many grams of MgSO4 are produced?
Remember:
• Step 1: Detemine the moles pf each reactant.
• Step 2: Determine the limiting reagent.
• Step 3: Determine the moles of the product.
• Step 4: Calculate the grams of the product.
Molar mass
AgBr
Solution
Step 1. (50 mL/1000)(5.00 M) = 0.2500 mol Li2SO4
(45 mL/1000)(4.50 M) = 0.2025 mol MgCl2
Step 2. Molar ratios for both reactants as well as the product are 1:1
so the smaller quantity of reactant (mol) = the quantity of product.
Smaller quantity of reactant = MgCl2 @ 0.2025 mol …
Step 3.
… ---> 0.2025 mol MgSO4.
Step 4. (0.2025 mol MgSO4)(24+32+16x4) = 24.3 g MgSO4.