Grade 11 Physical Sciences · Chemical Change

Identify the limiting reagent

Work out which reactant runs out first and use it to calculate how much product forms or how much of the other reactant is left.

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Limiting Reactant Grade 11: IntroductionKevinmathscience

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Question 5.6 · May/June 2025 · Paper 2 · National (DBE) · 2 marks

The reaction between magnesium carbonate pellets,

MgCO_{3}

(s), and EXCESS dilute hydrochloric acid, HCℓ(aq), is used to investigate the effect of concentration on the rate of a reaction.

MgCO_{3} (s) + 2 HC ℓ (aq) \to MgC ℓ_{2} (aq) + H_{2} O (ℓ) + CO_{2} (g)

The apparatus used for this investigation is shown below. Diagram for this question

5.6 How will the final mass of

CO_{2}

(g) produced in EXPERIMENT 2 compare to that of EXPERIMENT 3? Choose from MORE THAN, LESS THAN or THE SAME.

Give a reason for the answer.(2)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: identify the limiting reagent

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3. Question 2.3.2 · November 2025 · Paper 2 · National (DBE) · 5 marks
The letters A to G in the table below represent organic compounds.
2.3 Compound E, $C_{3} H_{8}$ (g), reacts with oxygen, $O_{2}$ (g), according to the balanced equation:

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O(g)$

Initially $8$ cm $^{3}$ of compound E and $50$ cm $^{3}$ of oxygen were injected into a container of adjustable volume and allowed to react, as shown below.

Answer the questions that follow.
2.3.2 Calculate the TOTAL volume of the GASES present in the container at the end of the reaction.(5)
Show the memo (5 ticks)Memo
1. ✓
  (a) Volume of O2 used (with volume of C3H8) $V (O_{2})_{used} = 5 V (C_{3} H_{8}) = (5) (8) = 40$ cm $^{3}$
  1 mark · Use the mole ratio
2. ✓
  (b) Volumes of CO2 and H2O produced $V (CO_{2})_{formed} = 3 V (C_{3} H_{8}) = (3) (8) = 24$ cm $^{3}$
  $V (H_{2} O)_{formed} = 4 V (C_{3} H_{8}) = (4) (8) = 32$ cm $^{3}$
  1 mark · Use the mole ratio
3. ✓
  (c) Volume of O2 remaining $V (O_{2})_{unused} = 50 - 40 = 10$ cm $^{3}$
  1 mark · Identify the limiting reagent
4. ✓
  (d) Addition of the three volumesVolume of gas in the container $= 10 + 24 + 32$
  1 mark · Identify the limiting reagent
5. ✓
  (e) Final answerTotal volume of gas in the container $= 66$ cm $^{3}$ (or $0, 066$ dm $^{3}$ )
  1 mark · Identify the limiting reagent
4. Question 1.6 · November 2025 · Paper 2 · Provincial · 2 marks
Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, for example 1.11 E.
1.6 Consider the balanced equation:

$Na_{2} CO_{3}$ (aq) + 2 HCl (aq) → 2 NaCl (aq) + $H_{2} O$ ( $ℓ$ ) + $CO_{2}$ (g)

1 mol of $Na_{2} CO_{3}$ reacts with 1 mol of HCl to yield 0,2 mol of $CO_{2}$ .

Which ONE of the following combinations is CORRECT regarding the reaction?(2)
1. A Limiting reagent: $Na_{2} CO_{3}$ ; % yield of $CO_{2}$ : 40%
2. B Limiting reagent: $Na_{2} CO_{3}$ ; % yield of $CO_{2}$ : 20%
3. C Limiting reagent: HCl; % yield of $CO_{2}$ : 40%
4. D Limiting reagent: HCl; % yield of $CO_{2}$ : 20%
Show the memo (1 tick)Memo
1. ✓
  Choose the correct option1 mol $Na_{2} CO_{3}$ needs 2 mol HCl, but only 1 mol HCl is present, so HCl is the limiting reagent.
  From HCl: $CO_{2}$ : HCl = 1 : 2, so the theoretical yield is $0, 5$ mol $CO_{2}$ .
  % yield $= \frac{0 , 2}{0 , 5} \times 100 = 40%$
  
  Answer: C
  2 marks · Identify the limiting reagent
5. Question 7.5 · November 2025 · Paper 2 · Provincial · 2 marks
Learners crushed eggshells to determine the percentage of calcium carbonate that are found in a $1, 5$ g sample. The crushed eggshells are allowed to react with EXCESS hydrochloric acid. The balanced equation for the reaction:

$CaCO_{3} (s) + 2 HCl (aq) \to CaCl_{2} (aq) + CO_{2} (g) + H_{2} O (ℓ)$

The calcium carbonate reacted completely in 5 minutes. The graph below shows the carbon dioxide gas that was collected at STP.

Calculate the:
7.5 How does the carbon dioxide gas produced after 8 min compare to the carbon dioxide gas produced at 5 min?

Write down only HIGHER THAN, SMALLER THAN or EQUAL TO.

Give a reason for the answer.(2)
Show the memo (2 ticks)Memo
1. ✓
  CompareEqual to
  1 mark · Identify the limiting reagent
2. ✓
  ReasonAfter calcium carbonate completely reacted (at 5 min) there was no more $CO_{2}$ produced.
  1 mark · Identify the limiting reagent
6. Question 1.5 · November 2024 · Paper 2 · Provincial · 2 marks
1.5 $10$ moles of zinc (Zn) are allowed to react with $5$ moles of hydrochloric acid (HC $ℓ$ ) in a test tube according to the balanced equation:

$Zn + 2 HC ℓ \to ZnC ℓ_{2} + H_{2}$

How many moles of hydrochloric acid (HC $ℓ$ ) and zinc chloride ( $ZnC ℓ_{2}$ ) is in the test tube after the completion of the reaction?(2)
1. A HC $ℓ$ : $2, 5$ mol; $ZnC ℓ_{2}$ : $2, 5$ mol
2. B HC $ℓ$ : $0$ mol; $ZnC ℓ_{2}$ : $2, 5$ mol
3. C HC $ℓ$ : $5$ mol; $ZnC ℓ_{2}$ : $10$ mol
4. D HC $ℓ$ : $0$ mol; $ZnC ℓ_{2}$ : $10$ mol
Show the memo (1 tick)Memo
1. ✓
  Find the limiting reactant and the moles after the reaction $5$ mol HC $ℓ$ needs only $\frac{5}{2} = 2, 5$ mol Zn, and $10$ mol Zn is available, so Zn is in excess and HC $ℓ$ is the limiting reactant. All the HC $ℓ$ is used up: $0$ mol left.
  $n (ZnC ℓ_{2}) = \frac{1}{2} n (HC ℓ) = \frac{1}{2} (5) = 2, 5$ mol
  
  Answer: B
  2 marks · Identify the limiting reagent

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