4.1.3 Stoichiometric Calculations and Mole Ratio

This topic explains how to use balanced chemical equations to perform stoichiometric calculations. It covers mole ratios and how to extract quantitative information from chemical equations.

Definition

Stoichiometry: The calculation of quantities of reactants and products involved in a chemical reaction using a balanced chemical equation.

Mole Ratio: The ratio of moles of any two substances in a balanced chemical equation, given by the coefficients.

Main Idea

A balanced chemical equation tells us the mole ratio in which reactants combine and products form. Using this mole ratio, we can calculate the amounts of reactants needed or products formed in a chemical reaction. The coefficients in a balanced equation represent the number of moles of each substance.

Core Concept

• The coefficients represent the number of moles of each substance.
• The mole ratio is the ratio of these coefficients.
• This ratio allows conversion between moles of one substance and moles of another.

Detailed Explanation

What Does a Balanced Chemical Equation Tell Us?

Consider the following reaction:

2H_2(g) + O_2(g) → 2H₂O_(g)

We get the following information from this chemical equation:

i. Mole Interpretation: - 2 moles of H₂ combine with 1 mole of O₂ to produce 2 moles of H₂O.

ii. Particle Interpretation: - 2 × 6.02 × 10²³ molecules of hydrogen react with 6.02 × 10²³ molecules of oxygen to produce 2 × 6.02 × 10²³ molecules of water vapour.

iii. Mass Interpretation: - 4 g (2 moles) of H₂ combine with 32 g (1 mole) of O₂ to produce 36 g (2 moles) of H₂O.

iv. Volume Interpretation (at STP): - At STP, 2 × 22.414 dm³ of H₂ combine with 1 × 22.414 dm³ of O₂ to produce 2 × 22.414 dm³ of H₂O vapour.

Mole Ratio

The coefficients in a balanced chemical equation give the mole ratio.

For the reaction: 2H₂ + O₂ → 2H₂O

The mole ratios are: 2 mol H₂1 mol O₂ or 1 mol O₂2 mol H₂

2 mol H₂2 mol H₂O or 2 mol H₂O2 mol H₂

Example 4.3: Volume of Hydrogen from Magnesium and HCl

Problem: When 100 g of magnesium is treated with dilute hydrochloric acid, what volume of hydrogen can be collected at STP?

Chemical Equation: Mg_(s) + 2HCl_(aq) → MgCl_2(aq) + H_2(g)

Solution:

From the equation: - 1 mol of Mg produces 1 mol of H₂ at STP.

Moles of Mg: Moles of Mg = 100 g24 g/mol = 4.17 mol

From the equation: - 1 mol of Mg produces 1 × 22.4 dm³ of H₂ at STP.

So, 4.17 mol Mg will produce: 22.4 dm³ × 4.17 = 93.408 dm³ of H₂ at STP

Final Answer: 93.408 dm³ of H₂ at STP.

Example 4.4: Dissolution of NaCl in Water

Chemical Equation: NaCl H₂O→ Na⁺_(aq) + Cl^-_(aq)

Information from the equation:

i. The chemical equation shows that 1 mole of NaCl when dissolved in water gives 1 mole of Na⁺ ions and 1 mole of Cl⁻ ions.

ii. When 6.02 × 10²³ formula units of NaCl are dissolved in water, they produce 6.02 × 10²³ Na⁺ and 6.02 × 10²³ Cl⁻ ions.

iii. 58.5 g of NaCl dissolved in water gives 23 g of Na⁺ ions and 35.5 g of Cl⁻ ions.

Intermediate Level

Information from a Balanced Equation

Mole Ratio

For the reaction: 2H₂ + O₂ → 2H₂O

Mole ratios: 2 mol H₂1 mol O₂ , 2 mol H₂O2 mol H₂ , 2 mol H₂O1 mol O₂

Example 4.3: Volume of H₂ from Mg and HCl

• Mass of Mg = 100 g
• Molar mass of Mg = 24 g/mol

Required: - Volume of H₂ at STP

Equation: Mg + 2HCl → MgCl₂ + H₂

• Moles of Mg = 10024 = 4.17 mol
• 1 mol Mg → 1 mol H₂
• 4.17 mol Mg → 4.17 mol H₂
• Volume of H₂ = 4.17 × 22.4 = 93.408 dm³

Final Answer: 93.408 dm³

Important Points

• 1. Balanced equation gives mole ratio.
• 2. Coefficients = moles of each substance.
• 3. Mole ratio converts between substances.
• 4. Mass = moles × molar mass.
• 5. Volume (STP) = moles × 22.414 dm³.
• 6. Particles = moles × 6.02 × 10²³.
• 7. Example: 2H₂ + O₂ → 2H₂O.

Remember

• - Balanced equation: Gives mole ratio
• - Coefficients: Moles
• - Mole ratio: Converts between substances
• - Mass: Moles × molar mass
• - Volume (STP): Moles × 22.414 dm³
• - Particles: Moles × 6.02 × 10²³

Common Mistakes

• 1. Forgetting to balance the equation first: Stoichiometry requires a balanced equation.
• 2. Mixing up mole ratios: Use coefficients correctly.
• 3. Forgetting to convert mass to moles: Always convert to moles first.
• 4. Using wrong molar mass: Check the periodic table.
• 5. Forgetting molar volume at STP: 22.414 dm³.

Quick Revision

• Balanced equation gives mole ratio.
• Coefficients = moles of each substance.
• Mole ratio converts between substances.
• Mass = moles × molar mass.
• Volume (STP) = moles × 22.414 dm³.
• Particles = moles × 6.02 × 10²³.
• Example: 2H₂ + O₂ → 2H₂O.

60-Second Revision

• Balanced equation: Gives mole ratio
• Coefficients: Moles
• Mole ratio: Converts between substances
• Mass: Moles × molar mass
• Volume (STP): Moles × 22.414 dm³
• Particles: Moles × 6.02 × 10²³

QUALITY VALIDATION

• Topic: 4.1.3 Stoichiometric Calculations and Mole Ratio
• Chapter: 4 (Stoichiometry)

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