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Chemistry · 05 / 10

Reactions: ratios and limiting reagent

High school · intermediate · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Units: numbers have dimensionsThe mole: connect mass and particlesBalance a reaction equation
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

A reaction equation conserves atoms and charge. Coefficients specify amount ratios, not mass ratios. Compare each available reactant amount divided by its coefficient: the smallest ratio sets maximum extent for a complete-reaction model.

n=0.6/2
THE PICTURE TO REMEMBERReactions: ratios and limiting reagent
  1. 01Balance
  2. 02Amounts in mol
  3. 03Compare n/coefficient
  4. 04Limit and products

WORKED EXAMPLE

From principle to reasoning

2 H₂ + O₂ → 2 H₂O. For 0.6 mol H₂ and 0.4 mol O₂, ratios are 0.3 and 0.4. H₂ limits; at most 0.6 mol water forms.

Find the limiting reactant

For 2H₂ + O₂ → 2H₂O, starting with 3 mol H₂ and 2 mol O₂ gives amount/coefficient ratios 1.5 and 2. The minimum sets maximum extent 1.5 mol: H₂ limits, 3 mol water forms and 0.5 mol O₂ remains in the complete-reaction model.

Solve a problem in this subject

  1. Identify species

    Write formulas and useful states. Distinguish atoms, ions, molecules and amount of substance. Total charge is conserved.

  2. Connect quantities

    Use m = nM and C = n/V with consistent units. V is the final solution volume.

  3. Use coefficients

    Balance before building an extent table. Compare initial amounts divided by coefficients to identify the limiting reactant.

  4. Check context

    Remaining amounts cannot be negative. Check conservation, units and assumptions: dilution, medium, completion or equilibrium.

YOUR TURN

01 · Try before checking

For 0.4 mol H₂ and 0.1 mol O₂, which reagent limits?

One hint

Compare amount divided by coefficient.

Reveal and understand the solution

Ratios are 0.2 for H₂ and 0.1 for O₂. O₂ limits and at most 0.2 mol water forms.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

In 2H₂ + O₂ → 2H₂O, how much O₂ reacts with 4 mol H₂?

A hint to get started

Read the 2:1 coefficient ratio.

Step-by-step solution
  1. Two moles H₂ consume one mole O₂.
  2. Four moles H₂ consume two moles O₂.
  3. Coefficients apply to amounts, not directly to masses.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

With 3 mol H₂ and 2 mol O₂, which reactant limits the preceding reaction?

A hint to get started

Compare initial amount divided by coefficient.

Step-by-step solution
  1. For H₂: 3/2 = 1.5; for O₂: 2/1 = 2.
  2. H₂ has the smaller ratio and is limiting.
  3. At most 2 × 1.5 = 3 mol H₂O forms.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

Theory predicts 10 g product but 8 g is recovered. Find yield.

A hint to get started

Compare recovered and theoretical masses.

Step-by-step solution
  1. Yield = 8/10.
  2. The yield is 0.8, or 80%.
  3. Losses or side reactions can lower recovered yield.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

The smallest raw mass or amount alone does not identify the limiting reagent.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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