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How much? The amount of chemical change : R2.1.3 Limiting and excess reactants IB DP Chemistry Study Notes

How much? The amount of chemical change : R2.1.3 Limiting and excess reactants IB DP Chemistry Study Notes - New Syllabus 2025

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

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Reactivity 2.1.3 – Limiting and excess reactants

Reactivity 2.1.3 – Limiting and excess reactants

Limiting Reactant

The limiting reactant (or limiting reagent) is the reactant in a chemical reaction that is completely consumed first. Once it is used up, the reaction cannot proceed further, and no additional product can form. It determines the maximum (theoretical) amount of product that can be produced.

  

Excess Reactant

The excess reactant is the substance that remains unreacted when the limiting reactant is fully consumed. It is present in a greater quantity than necessary based on the stoichiometric ratio.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from a given amount of limiting reactant, assuming complete reaction and no losses. It is calculated using the balanced chemical equation and the limiting reactant amount.

Experimental Yield

The experimental yield (or actual yield) is the quantity of product actually obtained from the reaction in a laboratory or industrial process. This is usually less than the theoretical yield due to factors such as incomplete reactions, side reactions, or loss of product during separation and purification.

Percentage Yield

Percentage yield is used to compare the efficiency of a reaction:

\( \text{Percentage Yield} = \left( \frac{\text{Experimental Yield}}{\text{Theoretical Yield}} \right) \times 100 \)

Why the Limiting Reactant Controls the Yield

In a chemical reaction, substances combine in fixed mole ratios determined by the balanced equation. If one reactant runs out before the others, the reaction must stop. The amount of product formed is limited by the amount of the limiting reactant.

How to Determine the Theoretical Yield

  1. Write a balanced chemical equation for the reaction.
  2. Convert the mass of each reactant into moles using \( n = \frac{m}{M} \).
  3. Compare the mole ratio from the equation to identify the limiting reactant.
  4. Use the mole ratio to calculate how many moles of product can form from the limiting reactant.
  5. Convert the moles of product to mass using \( m = n \times M \).

Steps to Identify Limiting and Excess Reactants

  1. Write a balanced chemical equation for the reaction.
  2. Convert the given masses of reactants into moles using:

    \( n = \frac{m}{M} \)

  3. Compare the mole ratio of the actual amounts with the theoretical mole ratio from the equation.
  4. Determine which reactant is limiting (the one that produces less product or is insufficient to fully react).

Example

Consider the reaction:

\( \text{2Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 \)

If 5.40 g of aluminum reacts with 6.35 g of chlorine gas, determine the limiting reactant and calculate the theoretical yield of aluminum chloride.

▶️Answer/Explanation
  • Molar mass of Al = 26.98 g/mol → \( n(\text{Al}) = \frac{5.40}{26.98} \approx 0.200 \) mol
  • Molar mass of \( \text{Cl}_2 = 70.90 \) g/mol → \( n(\text{Cl}_2) = \frac{6.35}{70.90} \approx 0.0896 \) mol
  • From the balanced equation: 2 mol Al reacts with 3 mol \( \text{Cl}_2 \)
  • To react with 0.200 mol Al, we need \( \frac{3}{2} \times 0.200 = 0.300 \) mol \( \text{Cl}_2 \), but we only have 0.0896 mol
  • Therefore, chlorine is the limiting reactant
  • Mole ratio: \( \text{Cl}_2 : \text{AlCl}_3 = 3:2 \), so:
    \( n(\text{AlCl}_3) = \frac{2}{3} \times 0.0896 \approx 0.0597 \) mol
  • Molar mass of \( \text{AlCl}_3 = 133.33 \) g/mol → mass = \( 0.0597 \times 133.33 \approx 7.96 \) g

Theoretical Yield = 7.96 g of \( \text{AlCl}_3 \)

Key Takeaways

  • The limiting reactant determines how much product can form.
  • The theoretical yield is calculated based on the limiting reactant and assumes no losses.
  • The experimental yield is what is actually obtained.
  • Percentage yield evaluates reaction efficiency.

Steps to Calculate Mass of Excess Reactant Remaining

  1. Use the limiting reactant to calculate how many moles of the excess reactant are actually needed.
  2. Subtract the moles used from the initial moles of excess reactant.
  3. Convert the remaining moles to grams using:

    \( m = n \times M \)

Example

Given the reaction:

\( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)

If 28.0 g of \( \text{N}_2 \) reacts with 6.00 g of \( \text{H}_2 \), determine the limiting reactant and calculate the mass of ammonia produced.

▶️Answer/Explanation
  • Molar mass of \( \text{N}_2 = 28.02 \) g/mol → \( n(\text{N}_2) = \frac{28.0}{28.02} \approx 1.00 \) mol
  • Molar mass of \( \text{H}_2 = 2.02 \) g/mol → \( n(\text{H}_2) = \frac{6.00}{2.02} \approx 2.97 \) mol
  • From the equation: 1 mol \( \text{N}_2 \) reacts with 3 mol \( \text{H}_2 \)
  • To fully react with 1.00 mol \( \text{N}_2 \), we need 3.00 mol \( \text{H}_2 \), but only 2.97 mol is available
  • Therefore, H2 is the limiting reactant, and \( \text{N}_2 \) is in excess
  • From the ratio \( \text{H}_2 : \text{NH}_3 = 3:2 \), we get:
    \( n(\text{NH}_3) = \frac{2}{3} \times 2.97 \approx 1.98 \) mol
  • Molar mass of \( \text{NH}_3 = 17.03 \) g/mol → mass = \( 1.98 \times 17.03 \approx 33.7 \) g

Theoretical Yield = 33.7 g of \( \text{NH}_3 \)

  • Used moles of \( \text{N}_2 = \frac{1}{3} \times 2.97 = 0.99 \) mol
  • Initial moles of \( \text{N}_2 = 1.00 \) mol
  • Remaining = \( 1.00 – 0.99 = 0.01 \) mol
  • Mass = \( 0.01 \times 28.02 = 0.28 \) g of \( \text{N}_2 \) remains

Key Point: Always use mole ratio, not mass ratio, to determine limiting reactant

Example

Consider the reaction:

\( \text{Ca}(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2(g) \)

If 4.00 g of calcium reacts with 7.30 g of hydrochloric acid, determine the limiting reactant and calculate the mass of hydrogen gas produced.

▶️Answer/Explanation
  • Step 1: Molar Masses
  • \( \text{M(Ca)} = 40.08 \) g/mol → \( n(\text{Ca}) = \frac{4.00}{40.08} \approx 0.0998 \) mol
  • \( \text{M(HCl)} = 1.01 + 35.45 = 36.46 \) g/mol → \( n(\text{HCl}) = \frac{7.30}{36.46} \approx 0.200 \) mol
  • Step 2: Mole Ratio
  • From the equation: \( \text{Ca} : \text{HCl} = 1:2 \)
  • To react with 0.0998 mol Ca, we need \( 0.0998 \times 2 = 0.1996 \) mol HCl
  • We have 0.200 mol HCl, which is slightly more than required
  • So, calcium is the limiting reactant
  • Step 3: Product Calculation
  • From the equation: \( \text{Ca} : \text{H}_2 = 1:1 \), so \( n(\text{H}_2) = 0.0998 \) mol
  • \( \text{M(H}_2) = 2.02 \) g/mol → mass = \( 0.0998 \times 2.02 \approx 0.201 \) g

Theoretical Yield = 0.201 g of \( \text{H}_2 \)

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