This set practises five skills: choosing the right mole ratio, repairing particle explanations, checking equations, linking observation to conclusion, and writing an exact sticking point. The questions get harder as you go. All numbers and situations are invented for practice.
Answer on paper first. Then open the answer and compare your reasoning, not only your result. Use Ar values: H = 1, C = 12, N = 14, O = 16, Mg = 24, Cl = 35.5, Ca = 40, Fe = 56.
Part A: Ratios and calculations
Q1. 2H₂ + O₂ → 2H₂O. What mass of water forms from 4.0 g of hydrogen?
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Moles of H₂ = 4.0 ÷ 2 = 2.0 mol. Ratio H₂ : H₂O is 2 : 2, so 1 : 1, giving 2.0 mol of water. Mr of H₂O = 18, so mass = 2.0 × 18 = 36 g.
Check: oxygen needed is 1.0 mol = 32 g. Reactants 4.0 + 32 = 36 g, which equals the product mass.
Q2. N₂ + 3H₂ → 2NH₃. A student says 1 mol of nitrogen gives 17 g of ammonia. Find the error and give the correct mass.
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The student used a 1 : 1 ratio. The equation gives N₂ : NH₃ = 1 : 2, so 1 mol of N₂ forms 2 mol of NH₃. Mr of NH₃ = 17, so mass = 2 × 17 = 34 g.
Check: hydrogen needed is 3 mol = 6 g, and nitrogen is 28 g. Reactants total 34 g.
Q3. CaCO₃ → CaO + CO₂. A sample of 10.0 g of calcium carbonate decomposes completely. Find the mass of each product and check your answers.
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Mr of CaCO₃ = 40 + 12 + 48 = 100. Moles = 10.0 ÷ 100 = 0.100 mol. All ratios are 1 : 1 : 1.
CaO: Mr = 56, mass = 0.100 × 56 = 5.6 g. CO₂: Mr = 44, mass = 0.100 × 44 = 4.4 g.
Check: 5.6 + 4.4 = 10.0 g, the same as the starting mass.
Q4. Mg + 2HCl → MgCl₂ + H₂. 1.20 g of magnesium reacts completely. Find (a) the moles of hydrogen, (b) the volume of hydrogen at room conditions, taking the molar volume as 24 dm³/mol, and (c) the volume of 1.00 mol/dm³ hydrochloric acid needed.
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Moles of Mg = 1.20 ÷ 24 = 0.0500 mol.
(a) Ratio Mg : H₂ = 1 : 1, so 0.0500 mol of H₂.
(b) Volume = 0.0500 × 24 = 1.20 dm³ (1200 cm³).
(c) Ratio Mg : HCl = 1 : 2, so HCl = 0.100 mol. Volume = 0.100 ÷ 1.00 = 0.100 dm³ (100 cm³).
Check of masses: MgCl₂ formed = 0.0500 × 95 = 4.75 g, H₂ = 0.100 g, total 4.85 g. Reactants: 1.20 + HCl (0.100 × 36.5 = 3.65 g) = 4.85 g.
Part B: Explanation repair
Q5. Repair this answer: “Increasing the pressure makes a reaction between gases faster.”
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At higher pressure the same gas particles are in a smaller volume, so they are closer together. They collide more often, so more successful collisions happen each second, and the rate increases. The particles do not have more energy per particle unless the temperature changes.
Q6. Repair this answer: “A catalyst makes the reaction faster.”
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A catalyst provides an alternative reaction pathway with a lower activation energy. A larger fraction of collisions between particles now has enough energy to react. So more successful collisions occur each second, and the rate increases. The catalyst is not used up.
Part C: Equation checks
Q7. A student balances Na + Cl₂ → NaCl by writing Na + Cl₂ → NaCl₂. Explain the error and give a correct balanced equation.
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The student changed a subscript, which changes the formula to a substance that does not form here. Only coefficients may change. Correct: 2Na + Cl₂ → 2NaCl.
Check: Na 2 and 2, Cl 2 and 2.
Q8. Balance Fe + Cl₂ → FeCl₃, then check it with an atom count.
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2Fe + 3Cl₂ → 2FeCl₃. Fe: 2 and 2. Cl: 3 × 2 = 6 and 2 × 3 = 6. Both sides match.
Q9. Is Mg + Ag⁺ → Mg²⁺ + Ag balanced? Correct it if not.
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Atoms match, but the charge is +1 on the left and +2 on the right. Not balanced. Two silver ions are needed: Mg + 2Ag⁺ → Mg²⁺ + 2Ag. Charge: +2 on both sides. Atoms: Ag 2 and 2.
Part D: Observation, model, conclusion and sticking points
Q10. A solution Z gives a white precipitate when dilute nitric acid and then silver nitrate solution are added. Write the observation, the model and a careful conclusion.
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Observation: a white precipitate formed. Model: silver ions combined with a negative ion in Z to form an insoluble silver compound, Ag⁺ + Cl⁻ → AgCl if the ion is chloride. Conclusion: Z is consistent with containing chloride ions. A further test would confirm this, since the observation alone does not exclude every other possibility.
Q11. A student writes: “I got Q4(c) wrong and don’t understand acids.” Rewrite this as an exact sticking point using the five-line template. Invent plausible details.
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One acceptable answer: (1) Question: volume of acid for 1.20 g Mg. (2) Last sure line: moles of Mg = 0.0500 mol. (3) First unsure line: I wrote HCl = 0.0500 mol. (4) Expected 0.100 mol, because the equation has 2HCl. (5) Guess: I used the coefficient of H₂ instead of HCl. Any answer that names the step, the expected value and a guess earns full credit.
If you got these wrong
| Error type | Go to |
|---|---|
| Used 1 : 1 or the wrong coefficients (Q1, Q2, Q3, Q4) | Why correct arithmetic can use the wrong ratio |
| Explained with a trend, or used vague verbs (Q5, Q6) | Repair a particle-level explanation |
| Changed a subscript, or skipped an element or a charge (Q7, Q8, Q9) | Check a balanced equation independently |
| Conclusion too strong, or model placed in the observation (Q10) | Connect observation, model and conclusion |
| Wrote a vague note instead of a line-sized one (Q11) | Prepare an exact sticking point |
Log each error in the mistake log and retest queue, then check a ratio with the mole and equation-ratio tutor or an equation with the equation balance reasoning trainer. The module page shows the full route.
If the same error returns after you have reread the lesson, a teacher in online one-to-one Chemistry tuition can build a short set of questions around that exact habit.