Balancing Act
What is the first step you should take when balancing a chemical equation?
Start with the most complex compound and balance one element or ion at a time.
What is the mole ratio in 2 H₂ + O₂ → 2 H₂O?
2:1:2
Coefficients represent relative mole quantities: 2 mol H₂ react with 1 mol O₂ to produce 2 mol H₂O.
What is the formula for molarity?
M = moles of solute / liters of solution
What equation relates concentration and volume in a dilution?
M₁V₁ = M₂V₂
What is the molar mass of CO₂?
44.01 g/mol
Add atomic masses: C = 12.01, O₂ = 32.00 → total = 44.01 g/mol. This is essential for converting grams ↔ moles in stoichiometry.
Balance this equation: H₂ + Cl₂ → HCl
1 H₂ + 1 Cl₂ → 2 HCl
Each H₂ gives 2 H atoms and each Cl₂ gives 2 Cl atoms, so two HCl molecules are formed to conserve both H and Cl atoms.
When 5.923 g Mg reacts with excess HCl, how many grams of H₂ are produced?
0.491 g H₂
From Mg + 2HCl → H₂ + MgCl₂. Converting 5.923 g Mg → mol Mg → mol H₂ → g H₂ gives 0.491 g (Worksheet 4B).
What is the molarity of a 500.0 mL solution containing 0.2984 mol sucrose?
0.597 M C₁₂H₂₂O₁₁
Convert 500.0 mL → 0.500 L. M = 0.2984 ÷ 0.500 = 0.5968 M (Worksheet 5A).
125 mL of 6.00 M Na₂B₄O₇ diluted to 500.0 mL—what is the new molarity?
1.50 M
M₂ = (M₁V₁)/V₂ = (6.00×125)/(500) = 1.50 M (Worksheet 5A).
What type of reaction is 2 KClO₃ → 2 KCl + 3 O₂?
Decomposition reaction
One compound breaks down into simpler products (Worksheet 4I).
Balance: C₃H₈ + O₂ → CO₂ + H₂O
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Balance C first (3 CO₂), then H (8 → 4 H₂O), then O (10 O atoms → 5 O₂ molecules).
In the reaction CO + 2H₂ → CH₃OH, if 15.6 g CO reacts with 3.47 g H₂, which is the limiting reactant?
CO
Calculating both possible CH₃OH masses shows CO produces 34.55 g while H₂ gives 55.39 g—CO limits the reaction (Worksheet 4E).
How many moles are in 125 mL of a 0.333 M (NH₄)₃PO₄ solution?
0.0416 mol (NH₄)₃PO₄
n = M × V → 0.333 × 0.125 L = 0.0416 mol (Worksheet 5C).
What volume of 6.00 M HCl makes 2.00 L of 1.00 M solution?
333 mL
M₁V₁ = M₂V₂ → (6.00)V₁ = (1.00)(2.00) → V₁ = 0.333 L = 333 mL (Worksheet 5C).
In the redox reaction Al + CuSO₄ → Al₂(SO₄)₃ + Cu, which element is oxidized?
Aluminum (Al)
Al goes from 0 to +3 oxidation state — it loses electrons (Worksheet 6IVG).
Balance this equation: NH₃ + O₂ → N₂ + H₂O
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
First balance nitrogen (4 NH₃ → 2 N₂), then hydrogen (12 H → 6 H₂O), then oxygen (3 O₂ gives 6 O atoms).
Define theoretical yield.
The maximum possible amount of product that can form from given reactants under ideal conditions.
Theoretical yield is found using stoichiometry and assumes all limiting reactant converts to product.
How many grams of CH₃COOH are in 25.0 mL of 0.839 M solution?
1.26 g CH₃COOH
n = 0.839 × 0.025 L = 0.0210 mol; mass = n × 60.052 g/mol = 1.26 g (Worksheet 5D).
25.0 mL of 1.00 M FeCl₃ diluted to 250.0 mL → new molarity?
0.100 M
M₂ = (1.00×25)/250 = 0.100 M (Worksheet 5E).
A student dilutes 50.0 mL of 2.00 M NaOH to 250.0 mL. What’s the final concentration?
0.400 M
M₁V₁ = M₂V₂ → (2.00)(50.0) = M₂(250.0) → M₂ = 0.400 M (Worksheet 5D).
Balance this combustion: C₆H₁₄ + O₂ → CO₂ + H₂O
2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O
Start with carbon (12 C → 12 CO₂), then hydrogen (28 H → 14 H₂O), then balance O (38 O → 19 O₂).
If the theoretical yield of SO₃ is 18.75 g and the experimental yield is 11.81 g, what is the percent yield?
62.99% (≈63%)
% yield = (actual ÷ theoretical) × 100 = (11.81 ÷ 18.75) × 100 = 62.99% ≈ 63% (Worksheet 4F).
A 0.250 M solution of CoCl₂ is prepared. How many grams of CoCl₂ are needed to make 1.00 L of this solution?
34.5 g CoCl₂
Use moles = M × L = 0.250 × 1.00 = 0.250 mol. Then multiply by molar mass (129.83 g/mol): 0.250 × 129.83 = 34.5 g. This reinforces both molarity and molar mass skills.
You have 500.0 mL of 1.000 M FeCl₃. A student pipets 25.00 mL into a 250.0 mL flask and fills to the mark. What is the new molarity?
0.100 M FeCl₃
M₁V₁ = M₂V₂ → (1.000×25.0) = M₂(250.0) → M₂ = 0.100 M. Shows proportional dilution of 1:10.
Why does soda fizz when you open it?
Because carbon dioxide gas escapes from the liquid as pressure is released.