This page holds eleven original questions in rising difficulty, moving from formulas and equations through amounts to explanations. Every answer is worked in full so you can compare your method, not just your result.
Attempt each question on paper first, then open the answer. Each topic also has its own practice set in its module, reached from the Chemistry learning guide, and the original mixed-practice builder can build a longer session.
Use these values where a question needs them. Ar: H = 1, C = 12, N = 14, O = 16, Mg = 24, Al = 27, Cl = 35.5, Ca = 40. Molar volume of a gas: 24 dm³/mol at room temperature and pressure.
How to use the set
- Work in one quiet block of 35 to 45 minutes.
- Write every step, including the ratio line and the units.
- Mark yourself honestly, then note the cause of each error: missing fact, particle idea, equation, calculation or fit to the question.
- Put each error in the mistake log.
Questions
Q1. Calculate the relative formula mass (Mr) of calcium hydroxide, Ca(OH)₂.
Show answer
Count the atoms: 1 Ca, 2 O and 2 H, because the subscript 2 outside the bracket applies to both O and H.
Mr = 40 + (2 × 16) + (2 × 1) = 40 + 32 + 2 = 74
Q2. Balance: Mg + HCl → MgCl₂ + H₂
Show answer
Chlorine: the right has 2, so put 2 in front of HCl. Then hydrogen is 2 on the left and 2 in H₂ on the right. Magnesium is 1 and 1.
Mg + 2HCl → MgCl₂ + H₂
Check: Mg 1 = 1, H 2 = 2, Cl 2 = 2.
Q3. Balance: Fe₂O₃ + CO → Fe + CO₂
Show answer
Iron: the left has 2, so put 2 in front of Fe. Oxygen: the left has 3 from Fe₂O₃, and carbon monoxide supplies 1 each, so try 3CO on the left and 3CO₂ on the right to keep carbon equal.
Oxygen check: left = 3 + 3 = 6, right = 3 × 2 = 6.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Q4. Explain, using particles, why a gas can be compressed into a smaller volume but a solid cannot.
Show answer
In a gas, the particles are far apart with large gaps between them, so a push can move them closer and the volume falls. In a solid, the particles are already touching in a fixed pattern, so there is almost no empty space to remove.
The particles themselves do not shrink in either case. Only the spacing between them changes.
Q5. How many moles are there in 11.0 g of carbon dioxide, CO₂?
Show answer
Mr of CO₂ = 12 + (2 × 16) = 44.
Moles = mass ÷ Mr = 11.0 ÷ 44 = 0.25 mol
Q6. Magnesium reacts with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen forms from 2.4 g of magnesium?
Show answer
n(Mg) = 2.4 ÷ 24 = 0.10 mol.
Ratio n(Mg) : n(H₂) = 1 : 1, so n(H₂) = 0.10 mol.
Volume = 0.10 × 24 = 2.4 dm³
Q7. What mass of sodium hydroxide, NaOH (Mr = 40), is in 50 cm³ of a 0.20 mol/dm³ solution?
Show answer
Convert the volume first: 50 cm³ ÷ 1000 = 0.050 dm³.
n = 0.050 × 0.20 = 0.0100 mol.
Mass = 0.0100 × 40 = 0.40 g
Q8. A student finds that 20.0 cm³ of 0.150 mol/dm³ hydrochloric acid neutralises 25.0 cm³ of sodium hydroxide solution (NaOH + HCl → NaCl + H₂O). Find the concentration of the sodium hydroxide.
Show answer
n(HCl) = 0.0200 × 0.150 = 0.00300 mol.
Ratio NaOH : HCl = 1 : 1, so n(NaOH) = 0.00300 mol.
Concentration = 0.00300 ÷ 0.0250 = 0.120 mol/dm³
Check: a larger volume of alkali than acid means the alkali must be the weaker solution. 0.120 is less than 0.150, which fits.
Q9. Burning 4.8 g of magnesium in oxygen (2Mg + O₂ → 2MgO) gives 7.2 g of magnesium oxide in a real experiment. Calculate the percentage yield.
Show answer
n(Mg) = 4.8 ÷ 24 = 0.20 mol. Ratio Mg : MgO = 1 : 1, so the maximum is 0.20 mol of MgO.
Mr of MgO = 40, so the theoretical mass = 0.20 × 40 = 8.0 g.
Percentage yield = (7.2 ÷ 8.0) × 100 = 90%
Q10. 0.30 mol of hydrogen and 0.10 mol of oxygen are mixed and react fully: 2H₂ + O₂ → 2H₂O. Which is the limiting reactant, and what mass of water forms? (Mr of H₂O = 18.)
Show answer
0.10 mol of O₂ needs 2 × 0.10 = 0.20 mol of H₂. There is 0.30 mol of H₂, so hydrogen is in excess and oxygen is the limiting reactant.
n(H₂O) = 2 × 0.10 = 0.20 mol. Mass = 0.20 × 18 = 3.6 g
0.10 mol of H₂ remains unreacted.
Q11. Marble chips and marble powder of the same mass react with the same volume of the same acid. Explain, using particles, why the powder reacts faster.
Show answer
The powder has a much larger surface area for the same mass, so more marble particles are exposed to the acid at once. That means more frequent collisions between acid particles and marble particles each second, and a faster rate.
The total amount of marble has not changed, so the final amount of product is the same. Only the rate differs.
If you got these wrong
Route each error type to the lesson that repairs it.
| Question | If it went wrong | Go to |
|---|---|---|
| Q1 | Forgot to multiply inside the bracket | Calculate a relative formula mass |
| Q2 | Changed a formula instead of adding a coefficient | Balance atoms without changing subscripts |
| Q3 | Balanced iron but not oxygen | Check a balanced equation independently |
| Q4 | Said the particles get smaller | Particle models and changes of state |
| Q5 | Multiplied instead of dividing | Convert mass to amount |
| Q6 | Used the wrong ratio or forgot the molar volume | Gas amount and volume |
| Q7 | Used 50 instead of 0.050 | Convert solution volume first |
| Q8 | Chose the wrong substance for the ratio | Use an equation ratio |
| Q9 | Divided by the wrong mass | Percentage yield |
| Q10 | Compared the moles without the ratio | Identify a limiting quantity |
| Q11 | Wrote “faster” without a particle reason | Powder and lumps |
If your arithmetic is right but the answers are still wrong, see using the wrong mole ratio. If your equations do not balance, see changing subscripts.
When does practice need a teacher?
Some students read every worked answer here and still feel that the next question will go wrong in a new way. That is the point where listening to your reasoning helps more than another worksheet. In online one-to-one Chemistry tuition, a teacher can ask you to explain a problem aloud and notice where a sound idea turns into a wrong line.
For your route and scope, see the study route and revision page.