IB Chemistry How to Solve: IB Chemistry – Stoichiometry & Mole Calculations (Gas Laws, Limiting Reactant)

How to Solve: IB Chemistry – Stoichiometry & Mole Calculations (Gas Laws, Limiting Reactant)

Complete Guide

Introduction

"Mastering stoichiometry and gas laws doesn’t just get you 10–15% of your IB Chemistry exam—it’s the key to predicting how much fuel a rocket needs, how much medicine a patient requires, or how much CO₂ a car emits. One wrong mole ratio, and your entire calculation collapses. Today, we’ll break it down into steps so simple, you’ll solve even the trickiest exam questions in under 5 minutes."

WHAT YOU NEED TO KNOW FIRST

Before starting, you must already understand:
1. Balanced chemical equations – Coefficients represent mole ratios.
2. Molar mass calculations – How to find grams per mole from the periodic table.
3. Basic algebra – Rearranging equations to solve for unknowns.

If you’re shaky on any of these, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas You MUST Know

1. Mole Calculations (n = number of moles)

Formula: [ n = \frac{m}{M} ] - n = number of moles (mol) - m = mass (g) - M = molar mass (g/mol) MEMORISE THIS – You’ll use it in every stoichiometry problem.

2. Gas Laws (Ideal Gas Equation)

Formula: [ PV = nRT ] - P = pressure (kPa) - V = volume (dm³) - n = number of moles (mol) - R = 8.31 J/mol·K (gas constant, given on IB data booklet) - T = temperature (in Kelvin! °C + 273 = K) MEMORISE THIS – But check the data booklet for R and STP values.

Shortcut for STP: [ n = \frac{V}{22.7} ] - V = volume in dm³ at STP - 22.7 dm³/mol = molar volume at STP (given on IB data booklet)

3. Mole Ratios (From Balanced Equations)

Example: [ 2H₂ + O₂ → 2H₂O ] - 2 mol H₂ reacts with 1 mol O₂ to produce 2 mol H₂O. - Mole ratio: 2:1:2

MEMORISE THIS: Always use coefficients from the balanced equation for mole ratios.

4. Limiting Reactant Steps

1. Calculate moles of each reactant.
2. Compare to the mole ratio from the balanced equation.
3. The reactant with less than required is the limiting reactant.
4. All calculations must use the limiting reactant’s moles.

STEP-BY-STEP METHOD

Step 1: Write the Balanced Equation

• If not given, balance the equation first.
• Example: 2H₂ + O₂ → 2H₂O

Step 2: Convert Given Quantities to Moles

• For solids/liquids: Use ( n = \frac{m}{M} )
• For gases at STP: Use ( n = \frac{V}{22.7} )
• For gases not at STP: Use ( PV = nRT )

Step 3: Identify the Limiting Reactant (If Needed)

• Compare actual moles to required moles from the balanced equation.
• The reactant that runs out first is the limiting reactant.

Step 4: Use Mole Ratios to Find Moles of Product

• Multiply moles of limiting reactant by the mole ratio (product/reactant).

Step 5: Convert Moles of Product to Required Units

• Mass? ( m = n \times M )
• Volume of gas at STP? ( V = n \times 22.7 )
• Volume of gas not at STP? Use ( PV = nRT )

WORKED EXAMPLES

Example 1 – Basic (No Limiting Reactant)

Question: What mass of water is produced when 4.0 g of hydrogen gas reacts with excess oxygen? [ 2H₂ + O₂ → 2H₂O ]

Step 1: Balanced equation already given. Step 2: Convert H₂ to moles. [ n(H₂) = \frac{m}{M} = \frac{4.0}{2.0} = 2.0 \text{ mol} ] (Molar mass of H₂ = 2.0 g/mol)

Step 3: No limiting reactant (O₂ is in excess). Step 4: Use mole ratio (2:2 → 1:1). [ n(H₂O) = n(H₂) \times \frac{2}{2} = 2.0 \text{ mol} ]

Step 5: Convert moles of H₂O to mass. [ m(H₂O) = n \times M = 2.0 \times 18.0 = 36.0 \text{ g} ]

Answer: 36.0 g of water is produced.

What we did and why: - We converted mass to moles first because stoichiometry works in moles, not grams. - The mole ratio told us how many moles of product form. - Finally, we converted back to grams because the question asked for mass.

Example 2 – Medium (Limiting Reactant)

Question: 10.0 g of hydrogen gas reacts with 50.0 g of oxygen gas. What mass of water is produced? [ 2H₂ + O₂ → 2H₂O ]

Step 1: Balanced equation already given. Step 2: Convert both reactants to moles. [ n(H₂) = \frac{10.0}{2.0} = 5.0 \text{ mol} ] [ n(O₂) = \frac{50.0}{32.0} = 1.5625 \text{ mol} ]

Step 3: Identify limiting reactant. - Required ratio (from equation): 2 mol H₂ : 1 mol O₂ - Actual ratio: 5.0 mol H₂ : 1.5625 mol O₂ - For 1.5625 mol O₂, we need: [ 1.5625 \times 2 = 3.125 \text{ mol H₂} ] - We have 5.0 mol H₂ (more than needed), so O₂ is limiting.

Step 4: Use limiting reactant (O₂) to find moles of H₂O. [ n(H₂O) = n(O₂) \times \frac{2}{1} = 1.5625 \times 2 = 3.125 \text{ mol} ]

Step 5: Convert moles of H₂O to mass. [ m(H₂O) = 3.125 \times 18.0 = 56.25 \text{ g} ]

Answer: 56.3 g of water is produced.

What we did and why: - We checked both reactants because neither was in excess. - The limiting reactant determines the maximum product. - Always compare actual moles to required moles from the balanced equation.

Example 3 – Exam-Style (Gas Laws + Limiting Reactant)

Question: A student reacts 5.0 dm³ of methane (CH₄) with 15.0 dm³ of oxygen at STP. What volume of CO₂ is produced? [ CH₄ + 2O₂ → CO₂ + 2H₂O ]

Step 1: Balanced equation already given. Step 2: Convert volumes to moles (at STP). [ n(CH₄) = \frac{5.0}{22.7} = 0.220 \text{ mol} ] [ n(O₂) = \frac{15.0}{22.7} = 0.661 \text{ mol} ]

Step 3: Identify limiting reactant. - Required ratio: 1 mol CH₄ : 2 mol O₂ - Actual ratio: 0.220 mol CH₄ : 0.661 mol O₂ - For 0.220 mol CH₄, we need: [ 0.220 \times 2 = 0.440 \text{ mol O₂} ] - We have 0.661 mol O₂ (more than needed), so CH₄ is limiting.

Step 4: Use limiting reactant (CH₄) to find moles of CO₂. [ n(CO₂) = n(CH₄) \times \frac{1}{1} = 0.220 \text{ mol} ]

Step 5: Convert moles of CO₂ to volume (at STP). [ V(CO₂) = n \times 22.7 = 0.220 \times 22.7 = 5.0 \text{ dm³} ]

Answer: 5.0 dm³ of CO₂ is produced.

What we did and why: - We used STP shortcut (22.7 dm³/mol) because the question specified STP. - The limiting reactant was CH₄, even though O₂ had a larger volume. - The mole ratio told us 1:1 for CH₄:CO₂, so the volume stayed the same.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is all you need to remember for stoichiometry and gas laws:
1. Always start with a balanced equation—no shortcuts.
2. Convert everything to moles first—mass to moles, volume to moles (22.7 dm³/mol at STP), or PV = nRT if not at STP.
3. Find the limiting reactant by comparing actual moles to the required ratio from the equation.
4. Use the limiting reactant’s moles to find product moles via the mole ratio.
5. Convert back to what the question asks—mass, volume, or concentration.
6. Double-check units—Kelvin for temperature, kPa for pressure, dm³ for volume.
7. If stuck, write down every step—examiners give partial credit.

You’ve got this. Now go ace that exam."