Reacting Masses And Limiting Reagent
Learn and apply Reacting Masses And Limiting Reagent in the published Chemistry course sequence.
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The core idea
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Reacting Masses and Limiting Reagent: Orientation
Limiting reagent questions are ratio questions. Your marks come from showing the chain clearly: balanced equation → moles → compare with coefficients → limiting reagent → product/leftover.
Build this on Mole and Avogadro Constant and keep the Stoichiometry hub open so unit conversions and mole logic stay coherent.
Definitions (Must Know)
A. Limiting reagent
The limiting reagent is the reactant that is used up first, so it limits the amount of product formed.
B. Theoretical yield
The theoretical yield is the maximum amount of product predicted by the balanced equation (assuming the limiting reagent reacts completely).
C. Excess reagent
An excess reagent is a reactant that is present in more than the stoichiometric amount, so some is left unreacted after the reaction finishes.
Detailed Explanations
A. Limiting reagent workflow
Because the balanced equation fixes the mole ratio, the limiting reagent is the reactant that cannot supply enough moles to satisfy that ratio.
- Write a balanced equation.
- Convert all reactant amounts to moles.
- Compare actual moles to required ratio (or compute product moles from each reactant and choose the smaller).
- Use limiting reactant to find product moles, then convert to mass / volume / concentration.
Mini example: Mg + 2HCl → MgCl₂ + H₂
- If n(Mg) = 0.100 but n(HCl) = 0.150, then 0.100 mol Mg would need 0.200 mol HCl → HCl is limiting.
B. Leftover reactant
If asked, calculate:
- moles used (from limiting reactant and ratio)
- moles left = moles initial − moles used
C. Fast method: compare n ÷ coefficient
For a reaction aA + bB → …:
- compute n(A)/a and n(B)/b
- the smaller value is limiting
Worked Examples
Modelled example 1
Identify the Limiting Reagent
Problem
Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) If 2.40 g magnesium reacts with 0.150 mol hydrochloric acid, identify the limiting reagent. Use Aᵣ(Mg) = 24.0.
Study the worked solution
Convert magnesium to amount
Method
Divide magnesium mass by its molar mass.Reason
The equation compares reacting amounts rather than masses.Working
n(Mg) = 2.40/24.0 = 0.100 molTest the acid requirement
Reason
The equation requires two moles of HCl for every mole of Mg.Working
n(HCl\ required) = 2(0.100) = 0.200 molCompare with the supply
Method
Compare 0.150 mol available with 0.200 mol required.Reason
The acid runs out before all magnesium can react.Working
0.150 < 0.200, so HCl is the limiting reagent.
Guided practice 2
Calculate Product Mass from the Limiting Reagent
Problem
Using the same reaction and amounts, calculate the mass of MgCl₂ formed. Use Aᵣ(Mg) = 24.0 and Aᵣ(Cl) = 35.5.
Try this before viewing the solution
Hints
Hint 1: start from the limiting reagent
Hint 2: apply the product ratio
View solution step by step
Find product amount
Method
Divide the limiting HCl amount by two.Reason
The balanced equation gives a 2:1 HCl-to-MgCl₂ ratio.Working
n(MgCl₂) = 0.150/2 = 0.0750 molFind molar mass
Reason
The product formula contains one Mg and two Cl atoms.Working
M(MgCl₂) = 24.0 + 2(35.5) = 95.0 g mol⁻¹Convert to mass
Method
Use m = nM.Reason
The calculated product amount is the maximum allowed by the limiting reagent.Working
m = (0.0750)(95.0) = 7.13 g
Common misconception 3
Correct a Direct-mass Comparison
Learner claim
For 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), 5.60 g iron is mixed with 4.80 g oxygen. A learner says oxygen is limiting because 4.80 g < 5.60 g. Identify the first error and determine the limiting reagent. Use Aᵣ(Fe) = 56.0 and M(O₂) = 32.0 g mol⁻¹.
Diagnose before deciding
View solution step by step
Convert both masses to amounts
Method
Calculate moles before comparing reactants.Reason
Equal masses do not represent equal particle amounts, and the equation uses a 4:3 mole ratio.Working
n(Fe) = 5.60/56.0 = 0.100 mol; n(O₂) = 4.80/32.0 = 0.150 mol.Normalise by coefficients
Method
Compare 0.100/4 with 0.150/3.Reason
The smaller reaction extent reaches zero first.Working
0.0250 < 0.0500, so iron is limiting.
Examiner practice 4
Calculate Unreacted Magnesium
Problem
In the original magnesium–hydrochloric acid mixture, calculate the mass of magnesium left unreacted. [4 marks]
Try this before viewing the solution
View solution step by step
Use the limiting acid
1 markMethod
Start with 0.150 mol HCl.Reason
The limiting reagent determines how much magnesium can react.Working
n(HCl) = 0.150 molFind magnesium used
1 markReason
Two moles of HCl react with one mole of Mg.Working
n(Mg\ used) = 0.150/2 = 0.0750 molFind magnesium left
1 markReason
Excess remaining equals initial amount minus amount consumed.Working
n(Mg\ left) = 0.100-0.0750 = 0.0250 molConvert to mass
1 markMethod
Multiply the remaining amount by 24.0 g mol⁻¹.Reason
The question asks for mass rather than amount.Working
m = (0.0250)(24.0) = 0.600 g
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Credit the ratio, subtraction and mass conversion separately.
Challenge 5
Analyse an Ammonia Batch
Problem
In a batch calculation, 14.0 kg nitrogen is mixed with 2.00 kg hydrogen: N₂(g) + 3H₂(g) → 2NH₃(g) Assuming complete reaction, identify the limiting reagent and calculate the theoretical ammonia mass and the mass of excess reactant left. Use M(N₂) = 28.0, M(H₂) = 2.00 and M(NH₃) = 17.0 g mol⁻¹.
Try this before viewing the solution
Hints
Hint 1: align mass units
Hint 2: compare reaction extents
View solution step by step
Convert both reactants to amounts
Method
Convert kilograms to grams, then divide by molar mass.Reason
Both reactants must be in the same amount basis before applying coefficients.Working
n(N₂) = 14000/28.0 = 500 mol; n(H₂) = 2000/2.00 = 1000 mol.Identify the limiting reagent
Reason
500/1 = 500 reaction units are available from nitrogen, but 1000/3 = 333 from hydrogen.Working
Hydrogen is limiting.Calculate ammonia mass
Reason
Three moles of hydrogen form two moles of ammonia.Working
n(NH₃) = (2/3)(1000) = 667 mol; m = (667)(17.0) = 1.13 × 10⁴ g = 11.3 kg.Calculate nitrogen left
Method
Subtract nitrogen consumed from nitrogen supplied.Reason
One mole of nitrogen reacts per three moles of limiting hydrogen.Working
n(N₂\ used) = 1000/3 = 333 mol; n(N₂\ left) = 167 mol; m = (167)(28.0) = 4.67 kg.
Mind Stretchers
Mind stretcher 1Extension
In the reaction 2CO + O₂ → 2CO₂, 0.80 mol of CO reacts with 0.30 mol of O₂. Find the moles of CO₂ formed.
Show Hint
Convert each reactant to moles and compare the available amount divided by its equation coefficient.
Show Answer
Mark scheme:
- Need 1 mol O2 per 2 mol CO. For 0.80 mol CO, required O2 = 0.80/2 = 0.40 mol.
- Available O2 = 0.30 mol, so O2 is limiting.
- 1 mol O2 gives 2 mol CO2, so CO2 formed = 2(0.30) = 0.60 mol.
Mind stretcher 2: Using gas mass to determine purityExtension
Question. A 10.0 g impure sample of CaCO₃ produces 1.76 g of CO₂ with excess acid. Calculate the percentage purity. Use M(CO₂) = 44.0 and M(CaCO₃) = 100 g mol⁻¹.
Show Hint
CaCO₃ and CO₂ are in a 1:1 mole ratio.
Show Answer
n(CO₂) = 1.76/44.0 = 0.0400 mol, so the sample contained 0.0400 mol or 4.00 g of CaCO₃. Purity = (4.00/10.0) × 100 = 40.0%.