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How much? The amount of chemical change : R2.1.2 Using the mole ratio IB DP Chemistry Study Notes

How much? The amount of chemical change : R2.1.2 Using the mole ratio IB DP Chemistry Study Notes - New Syllabus 2025

How much? The amount of chemical change  – IB DP Chemistry- Study Notes

IITian Academy excellent Introduction to the Particulate Nature of Matter – Study Notes and effective strategies will help you prepare for your IB DP Chemistry 2025 exam.

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Reactivity 2.1.2 – The Mole Ratio of an Equation

Reactivity 2.1.2 – The Mole Ratio of an Equation

Definition of Mole Ratio

The mole ratio refers to the fixed proportion of moles of reactants and products in a balanced chemical equation. It is essential in stoichiometric calculations and tells you how many moles of each substance are involved in the chemical reaction.

For example, in the reaction:

\( \text{2H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)

The mole ratio is \( \text{H}_2 : \text{O}_2 : \text{H}_2\text{O} = 2:1:2 \), meaning:

  • 2 moles of hydrogen react with 1 mole of oxygen
  • to form 2 moles of water

Avogadro’s Law

Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles.

Mathematically:

\( V \propto n \) or \( \frac{V}{n} = \text{constant} \)

  • \( V \) = volume of the gas in dm3
  • \( n \) = number of moles of the gas

STP – Standard Temperature and Pressure

  • Standard Temperature = 0°C (273.15 K)
  • Standard Pressure = 100 kPa (1 atm ≈ 101.3 kPa)
  • At STP, 1 mole of an ideal gas occupies 22.7 dm3

Key Formulae and Their Meanings

  • \( n = \frac{m}{M} \) – This formula is used to calculate the number of moles (n) from a known mass (m). Here:
    • \( n \) = number of moles (mol)
    • \( m \) = mass of substance (g)
    • \( M \) = molar mass of the substance (g/mol)
  • \( m = n \times M \) – Rearranged version of the above to calculate mass (m) when moles and molar mass are known.
  • \( V = n \times 22.7 \) – Used to calculate the volume of a gas (V) at STP. Here:
    • \( V \) = volume of gas (dm3)
    • \( n \) = number of moles of gas
    • 22.7 = molar volume at STP (dm3/mol)
  • \( C = \frac{n}{V} \) – Used to calculate the concentration (C) of a solution. Here:
    • \( C \) = concentration (mol/dm3)
    • \( n \) = number of moles of solute
    • \( V \) = volume of solution (dm3)

These equations form the foundation for performing calculations involving mole ratios, masses, gas volumes, and concentrations in both gaseous and solution-phase reactions.

Using Mole Ratios to Find Masses 

Step-by-step:

  1. Write a balanced chemical equation.
  2. Identify the known mass or moles.
  3. Use the mole ratio to calculate unknown moles.
  4. Convert moles to mass using molar mass.

Use the formula:

\( m = n \times M \)

Example

How many grams of oxygen are required to react with 4 moles of hydrogen?

\( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)

▶️Answer/Explanation
  • Mole ratio \( \text{H}_2 : \text{O}_2 = 2:1 \)
  • Oxygen needed = \( \frac{1}{2} \times 4 = 2 \) mol
  • Molar mass of \( \text{O}_2 = 32.00 \) g/mol
  • Mass = \( 2 \times 32 = 64 \) g

Using Mole Ratios to Find Gas Volumes

At STP, 1 mole of any ideal gas occupies 22.7 dm3. Use this value with the mole ratio to find volumes.

Example

What volume of oxygen is needed to react with 3 moles of hydrogen at STP?

\( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)

▶️Answer/Explanation
  • Ratio \( \text{H}_2 : \text{O}_2 = 2:1 \)
  • Oxygen = \( \frac{1}{2} \times 3 = 1.5 \) mol
  • Volume = \( 1.5 \times 22.7 = 34.05 \) dm3

Using Mole Ratios to Find Concentrations 

To find concentration, use \( C = \frac{n}{V} \). Volume must be in dm3 (1 dm3 = 1000 cm3).

Example

What is the concentration of 0.5 mol of NaOH in 0.5 dm3 of solution?

\( \text{NaOH}(aq) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l) \)

▶️Answer/Explanation
  • \( C = \frac{n}{V} = \frac{0.5}{0.5} = 1.00 \) mol/dm3

Additional Worked Examples

Example: Mass from a Decomposition Reaction

Calculate the mass of \( \text{CO}_2 \) produced when 10.00 g of \( \text{CaCO}_3 \) decomposes:

\( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \)

▶️Answer/Explanation
  • M(CaCO₃) = 100.09 g/mol
  • n = \( \frac{10.00}{100.09} \approx 0.100 \) mol
  • Ratio = 1:1 → n(\( \text{CO}_2 \)) = 0.100 mol
  • M(CO₂) = 44.01 g/mol → mass = \( 0.100 \times 44.01 = 4.40 \) g

Example: Volume of Gas from Reaction

What volume of \( \text{H}_2 \) is produced at STP when 2.00 g of zinc reacts with excess HCl?

\( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)

▶️Answer/Explanation
  • M(Zn) = 65.38 g/mol → \( n = \frac{2.00}{65.38} \approx 0.0306 \) mol
  • Ratio = 1:1 → n(\( \text{H}_2 \)) = 0.0306 mol
  • Volume = \( 0.0306 \times 22.7 = 0.695 \) dm3

Example: Mass from Concentration

25.0 cm3 of 1.00 mol/dm3 HCl reacts with NaOH. What is the mass of NaOH required?

\( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

▶️Answer/Explanation
  • V = 25.0 cm3 = 0.0250 dm3
  • n(HCl) = \( 1.00 \times 0.0250 = 0.0250 \) mol
  • Ratio = 1:1 → n(NaOH) = 0.0250 mol
  • M(NaOH) = 40.00 g/mol → mass = \( 0.0250 \times 40.00 = 1.00 \) g
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