The short answer
Moles, the n = m/M and n = cV shortcuts, mole ratios, and finding the limiting reagent.
Written and checked by GAMSAT tutors — not AI-generated.
Try the reasoning style
We treat forgetting as a failure — a lapse to be patched with reminders and records. Yet a mind that kept everything could not think; it would drown in the undifferentiated noise of every moment it had ever lived. To forget is not so much to lose information as to decide, mostly without our noticing, what was never worth keeping.
The author's argument relies most directly on which unstated assumption?
Pick an option to see how the tutor reasons to the answer — not just whether you were right.
Not quite — the answer is B.
Work backwards from the conclusion: a mind that ‘kept everything’ supposedly ‘could not think.’ That only follows if thinking means leaving most of experience out — so B is the premise the argument quietly rests on. A raises reliability, which the passage never weighs; C contradicts ‘mostly without our noticing’; D smuggles in a claim about intellect the passage never makes. The question rewards finding the hidden premise, not recalling a fact.
The mole is just a counting unit — like a "dozen," but of something. Almost every quantitative chemistry question is the same loop: convert to moles → use the balanced equation's ratio → convert back to mass, volume or concentration.
Two conversions do most of the work
Moles from mass: (M = molar mass, g/mol). Moles from solution: (c in mol/L, V in L). Learn these two and you can move between grams, litres and concentration freely.
The stoichiometry method
1. Balance the equation
The coefficients ARE the mole ratio. Nothing works until it's balanced.
2. Convert everything to moles
Use or to turn the given quantities into moles.
3. Find the limiting reagent
Divide each reactant's moles by its coefficient — the smallest value runs out first and controls the yield.
4. Use the ratio, then convert back
Scale to the product using the balanced ratio, then convert moles back to the unit asked for.
Worked example — limiting reagent
For , you mix 2 mol of with 3 mol of . Which runs out first, and how much forms?
Check yourself
How many moles are in 18 g of water (M = 18 g/mol)?
Check yourself
For 2 Al + 3 Cl₂ → 2 AlCl₃, you mix 4.0 mol of aluminium with 4.5 mol of chlorine. Which reactant runs out first, and how much AlCl₃ can form?
Check yourself
25.0 mL of 0.200 mol/L sodium hydroxide exactly neutralises a solution of sulfuric acid, by 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O. How many moles of H₂SO₄ were present?
Key takeaways
- A mole is a count: 6.022 × 10²³ particles.
- n = m/M (mass) and n = cV (solution) are the two conversions you reuse constantly.
- Balance first — the coefficients are the mole ratio.
- Limiting reagent = smallest (moles ÷ coefficient). It caps the product.
- Always finish by converting moles back to the unit the question asks for.
Practise this with real GAMSAT-style questions
A timed diagnostic, real practice questions with worked explanations, and your first essay marked on our rubric built from ACER's two published criteria.