General Chemistry: practice problems and exam questions

Moles and stoichiometry, acids and bases, gases and reaction rates: the calculations that come back in every first-year chemistry exam.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Stoichiometrymass to mass · limiting reactant · percent yieldcalculations
  2. Molarity & dilutionmolarity and mass · dilution · mixing solutionscalculations
  3. Ideal gas lawP from n, V, T · from mass · state change or densitycalculations
  4. Calorimetry & enthalpyheating with q = m·c·ΔT · calorimeter · enthalpy of formationcalculations
  5. Chemical equilibriumKc from concentrations · ICE table · ICE table with a quadraticcalculations
  6. pH & buffersstrong acids and bases · weak acids · bufferscalculations
  7. Reaction kineticsfirst order · conversion and second order · Arrheniuscalculations

Exam tips for General Chemistry

  • Convert everything to moles first; the balanced equation works in moles, not grams.
  • Find the limiting reactant before you calculate any product.
  • For weak acids, check whether the small-x approximation is valid.

Worked examples

StoichiometryEasy
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

For the reaction 2Mg + O2 → 2MgO, what mass m of MgO forms from 12 g of Mg? Assume the other reactants are in excess.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/mol

Find m (g)

Show hints
  1. Convert the given mass to moles: n = m/M.
  2. The coefficients give the mole ratio: 2 mol of product for every 2 mol of reactant.
  3. n(Mg) = 12/24.31 = 0.4936 mol.
Show the answer and the step-by-step solution

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Moles of reactant

n = 12/24.31 = 0.4936 mol

Mole ratio

n(MgO) = 0.4936·2/2 = 0.4936 mol

Mass of product

m = 0.4936·40.3 = 19.89 g

Answer

m = 19.89 g
Molarity & dilutionMedium
2 mol/LV1​ = ?0.5 mol/L1000 mL

You need 1000 mL of 0.5 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 2 mol/Lc2 = 0.5 mol/LV2 = 1000 mL

Find V1 (mL)

Show hints
  1. Dilution keeps the number of moles: c₁·V₁ = c₂·V₂.
  2. Solve for V₁ = c₂·V₂/c₁. You can keep the volumes in mL.
  3. c₂·V₂ = 500 mmol.
Show the answer and the step-by-step solution

Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.

Volume of stock solution

V1 = c2·V2/c1 = 0.5·1000/2 = 250 mL

Answer

V1 = 250 mL

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Ideal gas law calculatorMolarity calculatorDilution calculator (C1V1 = C2V2)pH calculatorSpecific heat calculator (q = mcΔT)Henderson–Hasselbalch buffer calculatorPercent yield calculatorStoichiometry calculator (grams to grams and limiting reactant)Molar mass calculatorCombined gas law calculator (Boyle, Charles and Gay-Lussac)Empirical formula calculator (and molecular formula)

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Practise this course for freePractise this courseTake a mock exam5 free problems every day. No card needed.