How much oxygen burns 4 mol of hydrogen?

The ratio H2 : O2 is 2 : 1, so n(O2) = 4 / 2 = 2 mol.

How much CaO comes from 100 g of calcium carbonate?

n = 100 / 100 = 1 mol; CaO = 1 x 56 = 56 g; CO2 = 1 x 44 = 44 g.

What mass of hydrogen comes from 12 g of magnesium?

Ratio 1 : 1, so n(H2) = 0.5 mol; V = 0.5 x 24 = 12 dm3.

S4 Chemistry: Mole and Stoichiometry Paper 3 (Equations, Limiting Reagent and Yield)

2026-11-03

S · 4

This paper works from coefficients: combustion, decomposition, metal with acid, the Haber process, then limiting reagent and percentage yield, closing on the unbalanced-equation slip.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Chemistry – Mole and Stoichiometry
Level	Challenge

02　Questions (10 items)

Foundation2H2 + O2 -> 2H2O. How many moles of oxygen are needed to burn 4 mol of hydrogen completely?
FoundationCaCO3 -> CaO + CO2. 100 g of calcium carbonate (100 g/mol) decomposes fully. What mass of CaO and of CO2 forms?
FoundationMg + 2HCl -> MgCl2 + H2. 0.5 mol of magnesium reacts fully: how many moles of hydrogen, and what volume at RTP?
Foundation2Al + 3Cl2 -> 2AlCl3. How many moles of chlorine are needed for 4 mol of aluminium?
IntermediateN2 + 3H2 -> 2NH3. 1 mol of nitrogen with 3 mol of hydrogen reacts fully: how many moles and what mass of ammonia (17 g/mol)?
Intermediate12 g of magnesium (24 g/mol) reacts with excess dilute hydrochloric acid: how many moles and what mass of hydrogen?
IntermediateCH4 + 2O2 -> CO2 + 2H2O. 0.2 mol of methane burns fully: what volume of oxygen at RTP, and what mass of CO2?
Challenge2H2 + O2 -> 2H2O. 2 mol of hydrogen is mixed with 0.5 mol of oxygen and ignited. Which is limiting, what mass of water forms, and how much hydrogen is left?
ChallengeA reaction has a theoretical yield of 20 g but gives 15 g. What is the percentage yield, and what mass would 90 percent yield give?
Error redoA student writes H2 + O2 -> H2O and calculates on a 1 : 1 ratio. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	The ratio H2 : O2 is 2 : 1, so n(O2) = 4 / 2 = 2 mol.	Read the coefficients; it is not 1 : 1.
2	n = 100 / 100 = 1 mol; CaO = 1 x 56 = 56 g; CO2 = 1 x 44 = 44 g.	Mass is conserved: 56 + 44 = 100 g, which matches.
3	Ratio 1 : 1, so n(H2) = 0.5 mol; V = 0.5 x 24 = 12 dm3.	Take moles from the coefficients, then multiply by 24.
4	Al : Cl2 = 2 : 3, so n(Cl2) = 4 x 3 / 2 = 6 mol.	Cross-multiply the ratio instead of answering 4.
5	N2 : NH3 = 1 : 2, so n(NH3) = 2 mol; m = 2 x 17 = 34 g.	The mole ratio is 1 : 2, not 1 : 1.
6	n(Mg) = 12 / 24 = 0.5 mol; n(H2) = 0.5 mol; m = 0.5 x 2 = 1 g.	H2 has molar mass 2; using 1 halves the answer.
7	n(O2) = 0.2 x 2 = 0.4 mol, V = 0.4 x 24 = 9.6 dm3; n(CO2) = 0.2 mol, m = 0.2 x 44 = 8.8 g.	Oxygen has coefficient 2, carbon dioxide 1; check both.
8	Hydrogen needs 0.5 x 2 = 1 mol and 2 mol is present, so oxygen limits; water = 0.5 x 2 = 1 mol = 18 g; hydrogen left = 2 – 1 = 1 mol.	The limiting reagent is the one in shortest supply relative to its coefficient.
9	Yield = 15 / 20 x 100 = 75 percent; at 90 percent, actual = 0.9 x 20 = 18 g.	Percentage yield is actual over theoretical, never the reverse.
10	Wrong: the equation is unbalanced. It should be 2H2 + O2 -> 2H2O, a 2 : 1 ratio, so 1 mol of oxygen needs 2 mol of hydrogen.	Using ratios on an unbalanced equation throws off every mole value.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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