AQA Combined Science (8464) 5.3.2.3–5.3.2.4 · Quantitative
Using moles — calculations and limiting reactants
Magnesium fizzes in acid until something runs out. Which one runs out, how much hydrogen do you get, and how do the masses tell you the balanced equation in the first place?
Combined Science · Higher tier
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The fizzing stops. Something ran out.
Magnesium fizzes in hydrochloric acid, giving off hydrogen gas: Mg + 2HCl → MgCl₂ + H₂. A student mixes 2.4 g of magnesium with 3.65 g of hydrogen chloride dissolved in water. After a while the fizzing stops.
If you had to guess, which ran out: the magnesium, or the hydrogen chloride?
When two substances react, the one that is completely used up is the limiting reactant: it limits the amount of product. The other is in excess, and some of it is left over. Chemists often use an excess of one reactant so that all of the other one reacts. To find the limiting reactant, convert each mass to moles (n = mass ÷ Mr). Then compare those moles with the ratio in the balanced equation, not with each other.
The reaction bench · each counter is 0.01 mol
Load the flask. Call which runs out. Then react.
0 of 2 mixes reacted
Magnesium 0.08 mol · Hydrogen chloride 0.12 mol
Commit first. Which runs out first?
Determine
Which reactant is limiting?
Each mix is magnesium and hydrogen chloride: Mg + 2HCl → MgCl₂ + H₂. Ar: Mg = 24. Mr: HCl = 36.5.
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The masses in a reaction also tell you its balanced equation. Convert each mass in grams to moles, then divide every amount by the smallest one. That gives the simplest whole-number ratio, and those numbers are the balancing numbers. Use each substance’s formula as written: hydrogen gas is H₂, so its Mr is 2, not 1; oxygen gas is O₂, Mr 32.
Think again
“The ratio came out as 1 : 1.5 : 1, so I rounded the 1.5 up to 2.”
Spot the flaw
BalanceIron reacts with chlorine: Fe + Cl₂ → FeCl₃. From the masses, the moles are Fe 0.10, Cl₂ 0.15 and FeCl₃ 0.10, a ratio of 1 : 1.5 : 1. What is the balanced equation?
Equations
Learn it · no chemistry equation sheetn = m ÷ Mr
amount (mol) = mass (g) ÷ relative formula mass
masses → moles → ÷ smallest
The simplest whole-number ratio gives the balancing numbers.
Command words in this lesson
- Balance
- Here, from masses: moles of every substance, divide by the smallest, whole numbers into the equation.
- Determine
- Reach the answer from the data given: which reactant runs out, from moles and the equation ratio.
- Calculate
- Show the working: moles, ratio, moles of product, then mass with its unit.
- Explain
- Link the limiting reactant to the amount of product, one step to the next.
Key fact
The limiting reactant is the one completely used up, and it fixes the amount of product. Find it from moles and the equation ratio, never from the masses alone.
Key note · AQA 5.3.2.3–5.3.2.4 (8464)
Using moles — calculations and limiting reactants
- The limiting reactant is completely used up; the other reactant is in excess.
- Limiting reactant: convert to moles → compare to equation ratio → whichever runs out first limits the amount of product.
- Balancing from masses: masses → moles → divide by the smallest → simplest whole-number ratio.
- n = mass ÷ Mr, using the formula as written: H₂ = 2, O₂ = 32, Cl₂ = 71.
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Not sure which reactant runs out first?
On the reaction bench each counter stands for 0.01 mol, and the reaction is shown one Mg + 2HCl → MgCl₂ + H₂ step at a time. In a real flask countless particles react at once, and the hydrogen bubbles away as a gas.