Year 10 Science | Victorian Curriculum 2.0
Chemical reactions & conservation of mass
Topic 04 | Chemical sciences | Practice

What you will learn

  • state the Law of Conservation of Mass,
  • write a word equation, then a symbol equation for a reaction,
  • balance chemical equations by inspecting atom counts,
  • use conservation of mass to calculate an unknown mass,
  • recognise the common reaction types: synthesis, decomposition, displacement, combustion.
Why does mass not change in a chemical reaction?

In 1789 Antoine Lavoisier weighed the sealed products of a reaction and showed the total mass was the same before and after — even though the appearance changed completely. Atoms are not created or destroyed in chemistry; they are only rearranged. Every correctly written equation must have the same number of atoms of each element on both sides. The 500500500-year-old alchemists’ dream of turning lead into gold fails chemically for this reason — only nuclear reactions change atoms themselves.

Where you'll see this
  • Fireworks: metal salts combust to give signature colours (Sr red, Cu blue-green, Na yellow).
  • Car engines: octane combusts with oxygen to produce CO2_22​ and water.
  • Baking: NaHCO3_33​ + acid →\to→ CO2_22​ gas makes bread rise.
  • Rust: iron + oxygen + water →\to→ hydrated iron oxide (mass of rust >>> mass of iron).
  • Photosynthesis: CO2_22​ + H2_22​O →\to→ glucose + O2_22​, driven by sunlight.
Worked example 0 Real-world example: why a rusted car weighs more

A 5 0005\,0005000 g steel plate is left in the rain. After three years it has corroded. Mass of rust collected: 5 3005\,3005300 g. Where did the extra 300300300 g come from? Does this violate the Law of Conservation of Mass?

  1. Rust is iron(III) oxide, written Fe2_22​O3⋅x_3 \cdot x3​⋅xH2_22​O. The iron combines with oxygen from the air and water from the rain.
  2. Reaction: 4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \to 2\text{Fe}_2\text{O}_34Fe+3O2​→2Fe2​O3​.
  3. The 300300300 g of extra mass is oxygen (and water) that joined the solid from the atmosphere.
  4. If you weighed the air + steel + water system as a whole, the total mass would be unchanged. Nothing was created — oxygen moved from gas into solid.

Key idea: mass is conserved across the whole system, including gases you cannot see. Apparent gains and losses come from gases entering or leaving.

1. Word and symbol equations

A word equation names reactants and products:

methane+oxygen→carbon dioxide+water.\text{methane} + \text{oxygen} \to \text{carbon dioxide} + \text{water}.methane+oxygen→carbon dioxide+water.

A symbol equation uses chemical formulas:

CH4+2O2→CO2+2H2O.\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}.CH4​+2O2​→CO2​+2H2​O.

Numbers in front of formulas are called coefficients. They multiply the whole formula, not just one element.

Worked example 1 Word to symbol

Convert this word equation to a symbol equation: “sodium + chlorine →\to→ sodium chloride”.

  1. Sodium is Na; chlorine gas is Cl2_22​; sodium chloride is NaCl.
  2. Unbalanced: Na+Cl2→NaCl\text{Na} + \text{Cl}_2 \to \text{NaCl}Na+Cl2​→NaCl.
  3. Count atoms: LHS has 111 Na, 222 Cl. RHS has 111 Na, 111 Cl. Not balanced.
  4. Balance Cl: put coefficient 222 in front of NaCl →\to→ Na+Cl2→2NaCl\text{Na} + \text{Cl}_2 \to 2\text{NaCl}Na+Cl2​→2NaCl.
  5. Now Na: LHS 111, RHS 222. Put 222 in front of Na: 2Na+Cl2→2NaCl2\text{Na} + \text{Cl}_2 \to 2\text{NaCl}2Na+Cl2​→2NaCl. Balanced.

2. Balancing chemical equations — step by step

Strategy:

  1. Write the unbalanced equation.
  2. Count atoms of each element on both sides.
  3. Balance one element at a time, leaving H and O until last.
  4. Use coefficients only — never change a formula (subscripts).
  5. Re-check every element at the end.
Worked example 2 Balancing methane combustion

Balance: CH4+O2→CO2+H2O\text{CH}_4 + \text{O}_2 \to \text{CO}_2 + \text{H}_2\text{O}CH4​+O2​→CO2​+H2​O.

  1. Carbon: 111 on each side — balanced.
  2. Hydrogen: LHS 444, RHS 222. Put 222 in front of H2_22​O: CH4+O2→CO2+2H2O\text{CH}_4 + \text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}CH4​+O2​→CO2​+2H2​O.
  3. Oxygen: LHS 222, RHS 2+2=42 + 2 = 42+2=4. Put 222 in front of O2_22​: CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}CH4​+2O2​→CO2​+2H2​O.
  4. Re-check: C 1=11 = 11=1; H 4=44 = 44=4; O 4=44 = 44=4. Balanced.
Worked example 3 Balancing a tricky one

Balance: Fe+O2→Fe2O3\text{Fe} + \text{O}_2 \to \text{Fe}_2\text{O}_3Fe+O2​→Fe2​O3​.

  1. Iron: LHS 111, RHS 222. Put 222 in front of Fe: 2Fe+O2→Fe2O32\text{Fe} + \text{O}_2 \to \text{Fe}_2\text{O}_32Fe+O2​→Fe2​O3​.
  2. Oxygen: LHS 222, RHS 333. Need a common multiple of 222 and 333: use 666.
  3. Put 333 in front of O2_22​ (gives 666) and 222 in front of Fe2_22​O3_33​ (gives 666 O): 2Fe+3O2→2Fe2O32\text{Fe} + 3\text{O}_2 \to 2\text{Fe}_2\text{O}_32Fe+3O2​→2Fe2​O3​.
  4. Now Fe: LHS 222, RHS 444. Adjust Fe: 4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \to 2\text{Fe}_2\text{O}_34Fe+3O2​→2Fe2​O3​.
  5. Check: Fe 4=44 = 44=4; O 6=66 = 66=6. Balanced.

3. Conservation of mass calculations

Because atoms are only rearranged, the total mass of reactants equals the total mass of products.

Worked example 4 Finding an unknown product mass

In the reaction 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}2H2​+O2​→2H2​O, 444 g of hydrogen reacts with 323232 g of oxygen. What mass of water is formed?

  1. Mass of reactants: 4+32=364 + 32 = 364+32=36 g.
  2. By conservation of mass, mass of products =36= 36=36 g.
  3. Mass of water formed =36= 36=36 g.
Worked example 5 Losing apparent mass to a gas

505050 g of calcium carbonate is heated: CaCO3→CaO+CO2\text{CaCO}_3 \to \text{CaO} + \text{CO}_2CaCO3​→CaO+CO2​. The solid remaining in the test tube has mass 282828 g. How much CO2_22​ gas escaped?

  1. Mass before: 505050 g.
  2. Mass after (solid): 282828 g.
  3. Mass lost as gas: 50−28=2250 - 28 = 2250−28=22 g of CO2_22​.
  4. This matches conservation: no mass is missing; it just left as invisible gas.

4. Types of chemical reactions

  • Synthesis (combination): A+B→ABA + B \to ABA+B→AB. E.g. 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}2H2​+O2​→2H2​O.
  • Decomposition: AB→A+BAB \to A + BAB→A+B. E.g. 2H2O2→2H2O+O22\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_22H2​O2​→2H2​O+O2​.
  • Displacement: a more reactive element pushes a less reactive one out. E.g. Zn+CuSO4→ZnSO4+Cu\text{Zn} + \text{CuSO}_4 \to \text{ZnSO}_4 + \text{Cu}Zn+CuSO4​→ZnSO4​+Cu.
  • Combustion: a fuel burns in oxygen to release CO2_22​ and water (if it contains C and H).
Worked example 6 Predicting a displacement product

Predict the products of Mg+FeSO4→\text{Mg} + \text{FeSO}_4 \toMg+FeSO4​→. Write a balanced equation.

  1. Magnesium is more reactive than iron, so Mg will displace Fe from the sulfate.
  2. Products: MgSO4_44​ and Fe metal.
  3. Equation: Mg+FeSO4→MgSO4+Fe\text{Mg} + \text{FeSO}_4 \to \text{MgSO}_4 + \text{Fe}Mg+FeSO4​→MgSO4​+Fe. Balanced as written.

Practice: Year 10

Fluency

Word and symbol equations

    1. State the Law of Conservation of Mass in one sentence.
    2. Write a word equation for the burning of hydrogen in oxygen.
    3. Write the symbol equation for the decomposition of water by electricity into hydrogen and oxygen gas.
    4. Identify the reactants and products in: Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2Zn+2HCl→ZnCl2​+H2​.
    5. What does the coefficient "222" mean in "2H2O2\text{H}_2\text{O}2H2​O"?
Fluency

Balancing

    Balance each equation.

    1. H2+Cl2→HCl\text{H}_2 + \text{Cl}_2 \to \text{HCl}H2​+Cl2​→HCl.
    2. Na+O2→Na2O\text{Na} + \text{O}_2 \to \text{Na}_2\text{O}Na+O2​→Na2​O.
    3. Al+O2→Al2O3\text{Al} + \text{O}_2 \to \text{Al}_2\text{O}_3Al+O2​→Al2​O3​.
    4. Mg+HCl→MgCl2+H2\text{Mg} + \text{HCl} \to \text{MgCl}_2 + \text{H}_2Mg+HCl→MgCl2​+H2​.
    5. C2H6+O2→CO2+H2O\text{C}_2\text{H}_6 + \text{O}_2 \to \text{CO}_2 + \text{H}_2\text{O}C2​H6​+O2​→CO2​+H2​O.
    6. Fe2O3+C→Fe+CO2\text{Fe}_2\text{O}_3 + \text{C} \to \text{Fe} + \text{CO}_2Fe2​O3​+C→Fe+CO2​.
Fluency

Conservation of mass

    1. 888 g of sulfur reacts completely with 888 g of oxygen. What is the mass of sulfur dioxide produced?
    2. 101010 g of limestone (CaCO3_33​) decomposes to give 5.65.65.6 g of calcium oxide. How much CO2_22​ was released?
    3. A 333 g magnesium ribbon burns in air and forms 555 g of magnesium oxide. How much oxygen reacted?
    4. 4.64.64.6 g of sodium combines with 3.553.553.55 g of chlorine gas. What mass of sodium chloride forms?
    5. 202020 g of iron rusts completely by combining with oxygen to form 28.628.628.6 g of iron(III) oxide. How much oxygen was used?
Reasoning

Reaction types and explain

    1. Classify each: (a) 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}2H2​+O2​→2H2​O; (b) CaCO3→CaO+CO2\text{CaCO}_3 \to \text{CaO} + \text{CO}_2CaCO3​→CaO+CO2​; (c) Cu+2AgNO3→Cu(NO3)2+2Ag\text{Cu} + 2\text{AgNO}_3 \to \text{Cu(NO}_3)_2 + 2\text{Ag}Cu+2AgNO3​→Cu(NO3​)2​+2Ag; (d) C3H8+5O2→3CO2+4H2O\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}C3​H8​+5O2​→3CO2​+4H2​O.
    2. A student tries to balance H2+O2→H2O2\text{H}_2 + \text{O}_2 \to \text{H}_2\text{O}_2H2​+O2​→H2​O2​ by changing the water formula. Why is this wrong?
    3. When iron rusts, the mass of the object appears to increase. Does this violate conservation of mass? Explain.
    4. A candle burns and loses mass. Explain where the mass went.
    5. Photosynthesis: 6CO2+6H2O→C6H12O6+6O26\text{CO}_2 + 6\text{H}_2\text{O} \to \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_26CO2​+6H2​O→C6​H12​O6​+6O2​. If 888888 g of CO2_22​ and 363636 g of water react, what is the combined mass of glucose and oxygen produced?
Problem solving

Real reactions

    1. When 555 g of copper carbonate decomposes, 3.23.23.2 g of copper oxide remains. What mass of CO2_22​ escaped? Write the balanced equation.
    2. A zinc strip placed in copper sulfate solution gradually becomes coated in copper. Write the balanced equation and classify the reaction.
    3. Write and balance the symbol equation for the complete combustion of butane (C4_44​H10_{10}10​) in oxygen.
    4. Sodium metal reacts vigorously with water to produce hydrogen gas and sodium hydroxide. Write the balanced equation.

Challenge

Reasoning

Harder reasoning

    1. 10.010.010.0 g of CaCO3_33​ is heated in an open crucible. The decomposition reaction produces CaO and CO2_22​. Explain how you could verify conservation of mass despite CO2_22​ escaping.
    2. A sealed glass ampoule contains 3.03.03.0 g of hydrogen and 24.024.024.0 g of oxygen. After ignition all the hydrogen is consumed. What is the mass of water formed, and what is left in the ampoule?
    3. Derive the balanced equation for the combustion of ethanol (C2_22​H5_55​OH) in oxygen, and use atom counting to verify the equation.
    4. Antoine Lavoisier weighed reactants and products in sealed vessels to prove conservation of mass. Why would open-vessel experiments of the day fail to reveal this law? Give two examples where an open vessel would give misleading mass readings.
Year 10 Science study companion | Practice