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StoichiometryChemistryHigh schoolabout 1:36

Grams of O2 from 24.5 g of KClO3: animated lesson

A narrated, step-by-step stoichiometry video lesson with captions. Final answer: 2KClO3 → 2KCl + 3O2; 24.5 g KClO3 → O2: 9.59587 g (0.29989 mol).

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.6 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    In the reaction KClO3 → KCl + O2, how many grams of O2 form from 24.5 g of KClO3?
  2. 2. Steps, part 1
    1. K:1c1=1c2Cl:1c1=1c2O:3c1=2c3\text{K}: 1c_{1} = 1c_{2} \\ \text{Cl}: 1c_{1} = 1c_{2} \\ \text{O}: 3c_{1} = 2c_{3}
      Write an atom balance for each element
    2. c1=2, c2=2, c3=3c_{1} = 2,\ c_{2} = 2,\ c_{3} = 3
      Solve the linear system with the smallest whole numbers
    3. 2KClO3→2KCl+3O22\mathrm{KClO_{3}} \rightarrow 2\mathrm{KCl} + 3\mathrm{O_{2}}
      Balanced equation
    4. n=24.5 g122.545 g/mol=0.199927 moln = \frac{24.5\ \text{g}}{122.545\ \text{g/mol}} = 0.199927\ \text{mol}
      Moles of KClO3
  3. 3. Steps, part 2
    1. n=0.199927×32=0.29989 mol,m=0.29989×31.998=9.59587 gn = 0.199927 \times \frac{3}{2} = 0.29989\ \text{mol},\quad m = 0.29989 \times 31.998 = 9.59587\ \text{g}
      O2 via the mole ratio 3 : 2
  4. 4. The answer
    2KClO3→2KCl+3O22\mathrm{KClO_{3}} \rightarrow 2\mathrm{KCl} + 3\mathrm{O_{2}}

    2KClO3 → 2KCl + 3O2

  5. 5. Check
    • K: 2 = 2, Cl: 2 = 2, O: 6 = 6 ✓
    • O2 is produced in a 3 : 2 mole ratio to KClO3.

Transcript

The problem. Here is the problem: In the reaction KClO3 → KCl plus O2, how many grams of O2 form from 24.5 g of KClO3? Balance atoms of every element, then use the coefficients as mole ratios.

Steps, part 1. Write an atom balance for each element. Give each species an unknown coefficient; atoms of every element must be equal on both sides. Solve the linear system with the smallest whole numbers. Exact elimination leaves one free coefficient; scaling by the least common denominator gives integers. Balanced equation. Moles of KClO3.

Steps, part 2. O2 via the mole ratio 3: 2.

The answer. So the answer is 2KClO3 → 2KCl plus 3O2; 24.5 g KClO3 → O2: 9.59587 g (0.29989 mol). Coefficients in a balanced equation count particles (moles), not grams. Convert grams to moles, apply the mole ratio, then convert back to grams with the other species prime molar mass. Assumes the reaction goes to completion with the known species fully consumed (or formed).

Check. K: 2 equals 2, Cl: 2 equals 2, O: 6 equals 6, which checks out. O2 is produced in a 3: 2 mole ratio to KClO3.

About stoichiometry

Use a balanced equation to find how many grams of product form from a given mass of reactant.

Grams of O2 from 24.5 g of KClO3: Stoichiometry Lesson