Solve a problem

StoichiometryChemistryHigh schoolabout 1:31

Grams of Al2O3 from 27 g of Al: animated lesson

A narrated, step-by-step stoichiometry video lesson with captions. Final answer: 4Al + 3O2 → 2Al2O3; 27 g Al → Al2O3: 51.0145 g (0.500334 mol).

SpeedLanguage

Chapters

Transcript

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. The problem
Worked example
In the reaction Al + O2 → Al2O3, how many grams of Al2O3 form from 27 g of Al?
2. Steps, part 1
1. $\text{Al}: 1c_{1} = 2c_{3} \\ \text{O}: 2c_{2} = 3c_{3}$
  Write an atom balance for each element
2. $c_{1} = 4,\ c_{2} = 3,\ c_{3} = 2$
  Solve the linear system with the smallest whole numbers
3. $4\mathrm{Al} + 3\mathrm{O_{2}} \rightarrow 2\mathrm{Al_{2}O_{3}}$
  Balanced equation
4. $n = \frac{27\ \text{g}}{26.982\ \text{g/mol}} = 1.00067\ \text{mol}$
  Moles of Al
3. Steps, part 2
1. $n = 1.00067 \times \frac{2}{4} = 0.500334\ \text{mol},\quad m = 0.500334 \times 101.961 = 51.0145\ \text{g}$
  Al2O3 via the mole ratio 2 : 4
4. The answer
$4\mathrm{Al} + 3\mathrm{O_{2}} \rightarrow 2\mathrm{Al_{2}O_{3}}$
4Al + 3O2 → 2Al2O3
5. Check
- Al: 4 = 4, O: 6 = 6 ✓
- Al2O3 is produced in a 2 : 4 mole ratio to Al.

Transcript

The problem. Here is the problem: In the reaction Al plus O2 → Al2O3, how many grams of Al2O3 form from 27 g of Al? 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 Al.

Steps, part 2. Al2O3 via the mole ratio 2: 4.

The answer. So the answer is 4Al plus 3O2 → 2Al2O3; 27 g Al → Al2O3: 51.0145 g (0.500334 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. Al: 4 equals 4, O: 6 equals 6, which checks out. Al2O3 is produced in a 2: 4 mole ratio to Al.

About stoichiometry

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

Grams of Al2O3 from 27 g of Al: Animated Step-by-Step Lesson