The problems are on the balancing equations worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A chemical equation shows reactants turning into products. These worked answers correct the 20 problems of the grade 8 science worksheet on balancing equations, from reactants and products to welding railway tracks: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the atom counts or the method first, and then the result in bold. Always recount your own balanced equation as a final check.
Level: Grade 8 science (Year 9 · Class 8) · Curriculum: MS-PS1-2, MS-PS1-5
Problem 1 Reactants and products
1. In addition, the reactants are copper and oxygen.2. Also, the product is copper oxide.
3. Yes, copper oxide is a new substance, so a chemical reaction has happened.
Problem 2 Signs of reaction
1. Also, chemical changes: (a), (d) and (e).2. Then, (a) a gas is formed (fizzing); (d) a colour change as rust forms; (e) heat and light given out.
Events a, d and e.
Problem 3 Counting atoms
1. Moreover, left: H 2, Cl 2.2. right: 2 × 1 H = 2, 2 × 1 Cl = 2.
3. Also, the counts are equal: it is balanced.
Problem 4 Balanced or not
(i) Left O 2, right O 1: not balanced.(ii) Na 2/2, Cl 2/2: balanced.
(iii) C 1/1, O 2/2: balanced.
Problem 5 Coefficient or subscript
1. Also, only the coefficient, the big number in front of a formula.2. H₂O₂ is a different substance (hydrogen peroxide), so the equation would no longer describe the reaction of making water.
Problem 6 Balance with a picture
1. Moreover, left: H 2, O 2. In addition, right: H 2, O 1.2. Make the oxygen even: 2 H₂O gives O 2 and H 4, so use 2 H₂: 2 H₂ + O₂ → 2 H₂O.
Problem 7 Balance three
1. 2 Na + Cl₂ → 2 NaCl2. Then, H₂ + Cl₂ → 2 HCl
3. Next, 2 Ca + O₂ → 2 CaO
Check: 1 Na 2/2, Cl 2/2; 2 H 2/2, Cl 2/2; 3 Ca 2/2, O 2/2.
Problem 8 Aluminium cooking pot
1. Also, left: Al 1, O 2. Also, right: Al 2, O 3.2. Moreover, the least common multiple of 2 and 3 is 6 oxygen atoms: 3 O₂ and 2 Al₂O₃, giving 4 Al: 4 Al + 3 O₂ → 2 Al₂O₃.
3. In addition, left Al 4, O 6; right Al 4, O 6.
Problem 9 Zinc in acid
1. Also, left: Zn 1, H 1, Cl 1. Also, right: Zn 1, Cl 2, H 2.2. Then, put 2 before HCl: Zn + 2 HCl → ZnCl₂ + H₂.
3. Moreover, hydrogen, H₂, is the gas.
Problem 10 Hydrogen peroxide
1. left: 2 × 2 = 4 H and 2 × 2 = 4 O. right: 2 × 2 = 4 H, and O is 2 + 2 = 4.2. Equal on both sides, so balanced.
3. Next, oxygen gas, O₂.
Problem 11 Baking soda in the oven
1. mass lost in total = 16.8 − 10.6 = 6.2 g. CO₂ = 6.2 − 1.8 = 4.4 g.2. Moreover, the conservation of mass: mass of products equals mass of reactants.
Problem 12 Open and closed flasks
1. 120.0 − 119.1 = 0.9 g.2. Nothing can leave a sealed flask, so the total mass stays the same.
3. In addition, the gas that escaped from the open flask carries the missing 0.9 g; atoms are conserved.
Problem 13 Spot the error
1. He changed a subscript, which creates the wrong substance (MgO₂ does not describe magnesium oxide).2. Also, change only coefficients: 2 Mg + O₂ → 2 MgO.
Problem 14 Word to symbol
1. 2 K + Cl₂ → 2 KCl2. Then, 2 Zn + O₂ → 2 ZnO
3. 2 × 2 = 4 atoms.
Problem 15 Check by tabulating
1. left: K 2, Cl 2, O 2 × 3 = 6. right: K 2, Cl 2, O 3 × 2 = 6.2. In addition, all three match: balanced.
Problem 16 Metals and chlorine
1. 2 Fe + 3 Cl₂ → 2 FeCl₃ (Fe 2/2, Cl 6/6).2. Then, 2 Al + 3 Cl₂ → 2 AlCl₃ (Al 2/2, Cl 6/6).
3. Next, 4 Na + O₂ → 2 Na₂O (Na 4/4, O 2/2).
Problem 17 Acids and metals
1. Mg + 2 HCl → MgCl₂ + H₂ (H 2/2, Cl 2/2).2. Then, 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂ (Al 2/2, H 6/6, Cl 6/6).
3. Next, Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O (H 4/4, O 2/2, Cl 2/2).
Problem 18 Fertiliser from air
1. 28 + 6 = 34 g on each side.2. 140 ÷ 28 = 5, so 5 × 34 = 170 g of ammonia.
3. 5 × 6 = 30 g of hydrogen.
Problem 19 Analysing sealed-flask data
1. Also, trial 1: −0.02 g; trial 2: +0.02 g; trial 3: −0.02 g; trial 4: −1.30 g.2. Then, trials 1, 2 and 3 agree within ±0.05 g.
3. Next, a leak let gas escape from the flask. Percentage lost = 1.30 ÷ 55.40 × 100 = 2.3 %.
Problem 20 Welding railway tracks
1. Fe 2/2, O 3/3, Al needs 2 on the left: Fe₂O₃ + 2 Al → Al₂O₃ + 2 Fe.2. 160 + 54 = 214 g; 102 + 112 = 214 g.
3. 80 g is half of 160 g, so iron = 56 g.
