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Year 12 in England

11th grade chemical reactions, acids and bases

7 11th grade chemical reactions, acids and bases skills, each with a worked example, free practice that checks every answer, and a printable worksheet. Start with a mixed review, or jump to the skill you need.

Mixed review

Ten questions drawn from every skill below.

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Moles and reacting masses

CCSS NGSS.HS-PS1-7England: KS4 Combined Science: Chemical changes (quantitative chemistry, Higher)

Example

How many moles are in 3.6 g of water (H₂O, Mr = 18)?

  1. 3.6 ÷ 18 = 0.2
  2. So the answer is: 0.2 mol

Answer: 0.2 mol

Builds on: Balancing chemical equations

Leads to: Count particles with the Avogadro constant, Volumes of gases from moles (24 dm³ at RTP), Concentration in mol/dm³, Percentage yield

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Count particles with the Avogadro constant

CCSS NGSS.HS-PS1-7England: GCSE Combined Science: Quantitative chemistry (Higher)

Example

How many oxygen molecules are in 0.5 mol? The Avogadro constant is 6.02 × 10²³ per mol. Give your answer as a number × 10²³ (to 2 decimal places).

  1. 0.5 × 6.02 = 3.01
  2. So 3.01 × 10²³
  3. So the answer is: 3.01

Answer: 3.01

Builds on: Moles and reacting masses

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Volumes of gases from moles (24 dm³ at RTP)

CCSS NGSS.HS-PS1-7England: GCSE Chemistry: Quantitative chemistry (volumes of gases, Higher)

Example

What volume does 1.5 mol of hydrogen gas occupy at room temperature and pressure, in dm³? (1 mol of any gas occupies 24 dm³ at RTP)

  1. 1.5 × 24 = 36
  2. So the answer is: 36 dm³

Answer: 36 dm³

Builds on: Moles and reacting masses

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Concentration in mol/dm³

CCSS NGSS.HS-PS1-7England: GCSE Chemistry: Quantitative chemistry (concentration of solutions, Higher)

Example

0.05 mol of hydrochloric acid is in 250 cm³ of solution. What is the concentration, in mol/dm³?

  1. 250 ÷ 1000 = 0.25 dm³
  2. 0.05 ÷ 0.25 = 0.2
  3. So the answer is: 0.2 mol/dm³

Answer: 0.2 mol/dm³

Builds on: Concentration in grams per dm³, Moles and reacting masses

Leads to: Titration calculations, Dilution calculations (c₁V₁ = c₂V₂)

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Percentage yield

CCSS NGSS.HS-PS1-7England: GCSE Chemistry: Quantitative chemistry (yield)

Example

A reaction could make 20 g of product in theory, but 9 g was actually collected. What is the percentage yield?

  1. 9 ÷ 20 × 100 = 45
  2. So the answer is: 45 %

Answer: 45 %

Builds on: Moles and reacting masses

Leads to: Find the limiting reactant and how much product forms

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Atom economy and greener reactions

CCSS NGSS.HS-PS1-7CCSS NGSS.HS-ETS1-3England: GCSE Chemistry: Quantitative chemistry (atom economy)

Example

CH₄ + H₂O → CO + 3H₂. The useful product is hydrogen (Mr 2). Relative masses of reactants: CH₄ = 16, H₂O = 18. What is the atom economy? Give 1 decimal place.

  1. useful product: 6; reactants: 34
  2. 6 ÷ 34 × 100 = 17.65
  3. So the answer is: 17.6 %

Answer: 17.6 %

Builds on: Relative formula mass

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Calculate energy changes from bond energies

CCSS NGSS.HS-PS1-4England: GCSE Combined Science: Energy changes (bond energies, Higher)

Example

CH₄ + 2O₂ → CO₂ + 2H₂O. Bond energies (kJ/mol): C–H 413, O=O 498, C=O 805, O–H 464. Calculate the energy change, in kJ/mol.

  1. broken: 4 × 413 + 2 × 498 = 2648
  2. made: 2 × 805 + 4 × 464 = 3466
  3. 2648 − 3466 = −818
  4. So the answer is: −818 kJ/mol

Answer: -818 kJ/mol

Builds on: Reaction profiles: energy change and activation energy

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11th grade Chemical reactions, acids and bases: Practice...