Writing and balancing chemical equations
A chemical equation shows reactants on the left and products on the right, with an arrow between. Because atoms are not made or destroyed, the number of each kind of atom must be the same on both sides. That is a balanced equation.
How to balance (hit and trial)
- Write correct formulas. Never change a formula (H₂O stays H₂O).
- Count atoms of each element on both sides.
- Put numbers (coefficients) in front of formulas, starting with the element that appears in the fewest formulas. Leave H and O for last.
- Check every element again. Use the smallest whole numbers.
- Add states: (s), (l), (g), (aq).
Example: CH₄ + O₂ → CO₂ + H₂O. C is balanced. H: 4 on left, so 2 H₂O. O: 2 + 2 = 4 on right, so 2 O₂. Result: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
Stoichiometric calculations
The coefficients tell moles, not grams. CH₄ + 2O₂ → CO₂ + 2H₂O means:
- 1 mol CH₄ reacts with 2 mol O₂ to give 1 mol CO₂ and 2 mol H₂O.
- In grams: 16 g + 64 g → 44 g + 36 g (total 80 g on each side).
- For gases at the same T and P, volumes follow the same ratio: 1 L + 2 L → 1 L + 2 L.
Four-step method
- Balance the equation.
- Change the given mass (or volume) into moles: n = m ÷ M.
- Use the mole ratio from the coefficients.
- Change moles of the wanted substance back into mass: m = n × M.
Limiting reagent
In real reactions the reactants are rarely in the exact recipe ratio. The limiting reagent is the reactant that is used up first. It stops the reaction and decides the amount of product. The other reactant is in excess; some of it is left over.
How to find it
- Change every reactant mass into moles.
- Divide each by its coefficient in the balanced equation.
- The smallest result is the limiting reagent.
- Calculate the product from the limiting reagent only.
- Leftover excess = amount taken − amount used.
Example: N₂ + 3H₂ → 2NH₃ with 50 kg N₂ and 10 kg H₂. Moles: N₂ = 50 000/28 = 1786 mol; H₂ = 10 000/2 = 5000 mol. Divide by coefficients: 1786/1 = 1786; 5000/3 = 1667. H₂ is smaller, so H₂ is limiting. NH₃ = 5000 × 2/3 = 3333 mol = 56.7 kg.
Reactions in solution: concentration terms
A solution has a solute (the smaller part, e.g. sugar) dissolved in a solvent (the bigger part, e.g. water). Concentration tells how much solute is present.
1. Mass percent (w/w)
Mass % = (mass of solute ÷ mass of solution) × 100. 2 g salt in 18 g water → 2/20 × 100 = 10%.
2. Mole fraction (x)
xA = moles of A ÷ total moles of all parts. Mole fractions add up to 1 and have no unit.
3. Molarity (M)
Molarity = moles of solute ÷ volume of solution in litres (mol L⁻¹). It changes slightly with temperature because liquids expand. For dilution: M₁V₁ = M₂V₂.
4. Molality (m)
Molality = moles of solute ÷ mass of solvent in kg (mol kg⁻¹). It does not change with temperature because mass does not change.
Try it at home
Stir 1 teaspoon (about 5 g) of salt into a glass (200 mL) of water. Work out its mass % and its molarity (NaCl M = 58.5): about 2.4% and 0.43 M. Then in the 3D free play, try 6 H₂ with 2 O₂ and predict which is limiting before you look.
Key formulas and definitions
- n = m / M; mole ratio = ratio of coefficients
- Limiting reagent: smallest (moles ÷ coefficient)
- Mass % = (mass of solute / mass of solution) × 100
- x_A = n_A / (n_A + n_B); x_A + x_B = 1
- Molarity M = moles of solute / volume of solution (L); M₁V₁ = M₂V₂
- Molality m = moles of solute / mass of solvent (kg)
Worked examples
1. Balance: Fe + O₂ → Fe₂O₃.
Line 1: Fe₂O₃ has 3 O; O₂ has 2. LCM is 6: put 2 before Fe₂O₃ and 3 before O₂. Line 2: Now 4 Fe on the right, so put 4 before Fe. Line 3: 4Fe + 3O₂ → 2Fe₂O₃. Check: Fe 4 = 4, O 6 = 6.
2. What mass of water forms when 16 g of methane burns completely? CH₄ + 2O₂ → CO₂ + 2H₂O.
Line 1: n(CH₄) = 16 ÷ 16 = 1 mol. Line 2: Ratio CH₄ : H₂O = 1 : 2 → 2 mol H₂O. Line 3: m = 2 × 18 = 36 g.
3. How many grams of oxygen are needed to burn 8 g of hydrogen to water?
Line 1: 2H₂ + O₂ → 2H₂O. Line 2: n(H₂) = 8 ÷ 2 = 4 mol. Line 3: n(O₂) = 4 × 1/2 = 2 mol. Line 4: m(O₂) = 2 × 32 = 64 g.
4. How much CaO forms by heating 50 g of CaCO₃? CaCO₃ → CaO + CO₂ (Ca = 40).
Line 1: M(CaCO₃) = 100; n = 50 ÷ 100 = 0.5 mol. Line 2: Ratio 1 : 1 → 0.5 mol CaO. Line 3: M(CaO) = 56 → m = 0.5 × 56 = 28 g.
5. 3 mol H₂ reacts with 2 mol O₂. Which is limiting? How much water forms and what is left?
Line 1: 2H₂ + O₂ → 2H₂O. Line 2: H₂: 3/2 = 1.5; O₂: 2/1 = 2. Smaller is H₂ → H₂ is limiting. Line 3: H₂O = 3 mol (ratio 1 : 1 with H₂). Line 4: O₂ used = 1.5 mol; left over = 2 − 1.5 = 0.5 mol O₂.
6. 50 kg N₂ and 10 kg H₂ react: N₂ + 3H₂ → 2NH₃. Find the limiting reagent and the mass of NH₃.
Line 1: n(N₂) = 50 000/28 = 1785.7 mol; n(H₂) = 10 000/2 = 5000 mol. Line 2: ÷ coefficients: N₂ 1785.7; H₂ 5000/3 = 1666.7 → H₂ limiting. Line 3: NH₃ = 5000 × 2/3 = 3333.3 mol. Line 4: m = 3333.3 × 17 = 56 667 g ≈ 56.7 kg.
7. 4 g of NaOH is dissolved in water to make 250 mL of solution. Find its molarity.
Line 1: M(NaOH) = 23 + 16 + 1 = 40; n = 4 ÷ 40 = 0.1 mol. Line 2: V = 250 mL = 0.250 L. Line 3: Molarity = 0.1 ÷ 0.250 = 0.4 M.
8. A solution has 1 mol of ethanol in 1 kg of water. Find molality and the mole fraction of ethanol.
Line 1: Molality = 1 mol ÷ 1 kg = 1 mol kg⁻¹. Line 2: n(water) = 1000 ÷ 18 = 55.6 mol. Line 3: x(ethanol) = 1 ÷ (1 + 55.6) = 0.0177.
9. How much water must be added to 100 mL of 2 M HCl to make it 0.5 M?
Line 1: M₁V₁ = M₂V₂ → 2 × 100 = 0.5 × V₂. Line 2: V₂ = 400 mL. Line 3: Water to add = 400 − 100 = 300 mL.
Common mistakes
- Changing subscripts to balance (writing H₂O₂ instead of 2H₂O). Only change the coefficients in front.
- Using the mass ratio as the mole ratio. Coefficients give moles; change grams to moles first.
- Picking the reactant with the smaller mass as limiting. Compare moles divided by coefficients, not grams.
- Using volume of solvent for molarity, or mass of solution for molality. Molarity uses litres of solution; molality uses kg of solvent.