The World of Metals and Non-metals – Numerical Problems with Stepwise Solutions
Class 7
Science — Chapter 4: The World of Metals and Non-metals
20 Numerical Problems with stepwise solutions — NCERT-aligned for CBSE Class 7.
CBSE Board Examination
Focus: simple stoichiometry, mass changes in reactions, alloys, corrosion calculations.
Content Bank — Important formulas & concepts
moles = mass / molar massmass = moles × molar mass- Use stoichiometric ratios from balanced equations (e.g.,
2Mg + O₂ → 2MgO) to relate moles. - Percent composition:
mass part / total mass × 100 - Rust approximation (using
Fe₂O₃): 4Fe + 3O₂ → 2Fe₂O₃ (molar mass Fe = 55.85, Fe₂O₃ = 159.70 g)
Numerical Problems & Stepwise Solutions (Topic-wise)
A. Reactions with Oxygen (Mg burning, formation of oxides)
Problem 1: If 24.305 g of magnesium burns completely in oxygen, what mass of magnesium oxide (MgO) will be formed? (At. masses: Mg = 24.305, O = 16.00)
Step 1: Calculate moles of Mg = 24.305 / 24.305 = 1 mol.
Step 2: Reaction: Mg + ½O₂ → MgO. 1 mol Mg → 1 mol MgO.
Step 3: Molar mass MgO = 24.305 + 16.00 = 40.305 g/mol.
Answer: Mass of MgO = 1 × 40.305 = 40.305 g.
Problem 2: If 48.61 g of magnesium is burned, what mass of MgO is produced?
Step 1: Moles Mg = 48.61 / 24.305 = 2.000 mol.
Step 2: Moles MgO = 2.000 mol. Mass = 2.000 × 40.305 = 80.610 g.
Answer: 80.610 g MgO.
B. Metals with Acids (hydrogen gas production)
Problem 3: 65.38 g of zinc reacts with excess dilute hydrochloric acid. How many grams of hydrogen gas are produced? (At. mass H = 1.008, H₂ = 2.016)
Step 1: Moles Zn = 65.38 / 65.38 = 1.000 mol.
Step 2: Reaction: Zn + 2HCl → ZnCl₂ + H₂. 1 mol Zn → 1 mol H₂.
Step 3: Mass H₂ = 1.000 × 2.016 = 2.016 g.
Answer: 2.016 g of hydrogen gas.
Problem 4: If 4.032 g of hydrogen gas is collected when zinc reacts with acid, how many grams of zinc reacted?
Step 1: Moles H₂ = 4.032 / 2.016 = 2.000 mol.
Step 2: Stoichiometry: 1 mol Zn → 1 mol H₂ ⇒ moles Zn reacted = 2.000 mol.
Step 3: Mass Zn = 2.000 × 65.38 = 130.76 g.
Answer: 130.76 g of zinc reacted.
C. Rusting / Corrosion (Iron & iron oxide calculations)
Problem 5: 55.85 g of iron (Fe) completely reacts to form iron(III) oxide (Fe₂O₃). What is the mass of Fe₂O₃ formed and how much mass is gained by iron (from oxygen)? (Use Fe = 55.85, O = 16.00)
Step 1: Molar mass Fe₂O₃ = 2×55.85 + 3×16.00 = 159.70 g/mol.
Step 2: Equation (simplified): 4Fe + 3O₂ → 2Fe₂O₃. So 4 mol Fe → 2 mol Fe₂O₃ ⇒ 1 mol Fe → 0.5 mol Fe₂O₃.
Step 3: Moles Fe = 55.85 / 55.85 = 1.000 mol ⇒ moles Fe₂O₃ = 0.5 mol.
Step 4: Mass Fe₂O₃ = 0.5 × 159.70 = 79.850 g.
Step 5: Mass gain = 79.850 - 55.85 = 24.000 g (oxygen combined).
Answer: 79.850 g Fe₂O₃ formed; iron gains 24.000 g from oxygen.
Problem 6: If 10.0 g of iron rusts (forms Fe₂O₃), approximately how much mass increase (due to oxygen) occurs?
Step 1: From Problem 5, 55.85 g Fe → mass increase 24.000 g. So per g Fe, increase = 24.000 / 55.85 = 0.429722... g per g Fe.
Step 2: For 10.0 g Fe, increase = 10.0 × 0.429722... ≈ 4.297 g.
Answer: Mass increases by about 4.297 g.
Problem 7: If 15.0 g of zinc is used as a sacrificial coating, how many grams of iron could it theoretically protect (1 mol Zn protects 1 mol Fe)? (Use Zn = 65.38, Fe = 55.85)
Step 1: 1 mol Zn (65.38 g) protects 1 mol Fe (55.85 g). So mass ratio Fe/Zn = 55.85 / 65.38 = 0.854237 (approx).
Step 2: For 15.0 g Zn, Fe protected = 15.0 × 0.854237 ≈ 12.814 g (rounded to three decimals).
Answer: Approximately 12.814 g of iron can be sacrificially protected by 15.0 g Zn.
D. Alloys & Percent Composition
Problem 8: Brass is 70% copper and 30% zinc by mass. In 200 g of brass, calculate the mass of copper and zinc.
Step 1: Copper mass = 200 × 0.70 = 140.00 g.
Step 2: Zinc mass = 200 × 0.30 = 60.00 g.
Answer: 140.00 g Cu and 60.00 g Zn.
Problem 9: Steel contains 2% carbon by mass. How much carbon is present in 50.0 kg of steel? (Give answer in grams.)
Step 1: Convert 50.0 kg to grams: 50.0 kg = 50,000 g.
Step 2: Carbon mass = 50,000 × 0.02 = 1000 g.
Answer: 1000 g (1 kg) of carbon.
Problem 10: How many moles of carbon are present in 12.0 g of carbon? (At. mass C ≈ 12.01 g/mol)
Step 1: Moles = mass / molar mass = 12.0 / 12.01 ≈ 0.999 mol (rounded to three decimals).
Answer: ≈ 0.999 mol of carbon.
E. Practical—Small mass change problems
Problem 11: If 10.0 g of magnesium reacts with oxygen, what mass of magnesium oxide is produced? (Use Mg = 24.305)
Step 1: Moles Mg = 10.0 / 24.305 ≈ 0.411438 mol.
Step 2: Molar mass MgO = 40.305 g/mol ⇒ mass MgO = 0.411438 × 40.305 ≈ 16.583 g.
Answer: ≈ 16.583 g MgO.
Problem 12: If 5.00 g of iron rusts to form Fe₂O₃, what is the mass of rust formed and the oxygen added?
Step 1: Moles Fe = 5.00 / 55.85 ≈ 0.08949 mol.
Step 2: Moles Fe₂O₃ = 0.08949 × 0.5 ≈ 0.044745 mol.
Step 3: Mass Fe₂O₃ = 0.044745 × 159.70 ≈ 7.149 g. Oxygen added = 7.149 − 5.00 = 2.149 g.
Answer: Fe₂O₃ ≈ 7.149 g; oxygen combined ≈ 2.149 g.
Problem 13: If 0.50 mol of zinc reacts with acid, what mass of hydrogen gas is produced?
Step 1: Moles H₂ = 0.50 mol (1:1 ratio with Zn) ⇒ mass H₂ = 0.50 × 2.016 = 1.008 g.
Answer: 1.008 g of H₂.
F. Larger-scale & proportion problems
Problem 14: If 100 g of iron rusts (forms Fe₂O₃), calculate the mass of oxygen that has combined with iron.
Step 1: From Problem 5 ratio: 55.85 g Fe → oxygen mass 24.000 g ⇒ per g Fe oxygen = 24.000 / 55.85 = 0.429722 g.
Step 2: For 100 g Fe, oxygen = 100 × 0.429722 ≈ 42.97 g (rounded to two decimals).
Answer: ≈ 42.97 g of oxygen combined.
Problem 15: An aluminum alloy of mass 250 g contains 10% copper. What is the mass of copper present?
Step 1: Copper mass = 250 × 0.10 = 25.00 g.
Answer: 25.00 g of copper.
Problem 16: How many grams of oxygen are needed to react completely with 2.00 mol of magnesium according to 2Mg + O₂ → 2MgO?
Step 1: 2 mol Mg require 1 mol O₂ ⇒ for 2.00 mol Mg, O₂ needed = 1.00 mol.
Step 2: Mass O₂ = 1.00 × 32.00 = 32.00 g.
Answer: 32.00 g of O₂.
Problem 17: If 3.00 mol of iron undergoes complete oxidation to Fe₂O₃, what mass of Fe₂O₃ is formed?
Step 1: 1 mol Fe → 0.5 mol Fe₂O₃ ⇒ 3.00 mol Fe → 1.50 mol Fe₂O₃.
Step 2: Mass = 1.50 × 159.70 = 239.55 g.
Answer: 239.55 g Fe₂O₃.
Problem 18: A 500 g brass object contains copper and zinc in 70:30 ratio. Calculate mass of copper and zinc.
Step 1: Copper = 500 × 0.70 = 350.00 g. Zinc = 500 × 0.30 = 150.00 g.
Answer: 350.00 g Cu and 150.00 g Zn.
Problem 19: A copper wire of mass 20.0 g loses 5.00% of its mass due to corrosion. What is the mass lost and the remaining mass?
Step 1: Mass lost = 20.0 × 0.05 = 1.000 g.
Step 2: Remaining mass = 20.0 − 1.000 = 19.000 g.
Answer: Lost = 1.000 g; Remaining = 19.000 g.
Problem 20: If 10.0 g of iron forms Fe₂O₃, what fraction of the product's mass is oxygen?
Step 1: From calculations: Fe₂O₃ mass formed ≈ 14.297 g when starting with 10.0 g Fe; oxygen mass combined ≈ 4.297 g.
Step 2: Fraction oxygen = 4.297 / 14.297 ≈ 0.301 (≈ 30.1%).
Answer: Oxygen is about 30.1% of the rust mass.
These problems use simple stoichiometry and percentage calculations appropriate to NCERT-level examples. Molar masses used: Mg = 24.305, Fe = 55.85, Zn = 65.38, O = 16.00, H₂ = 2.016, Fe₂O₃ = 159.70 g/mol.