20 Numerical Problems — Topic-wise (with stepwise solutions)

1. Reaction stoichiometry — displacement
   
   Balanced: CuSO₄ + Fe → FeSO₄ + Cu. If 15.9 g Cu is produced, how many grams of Fe reacted? (M_Cu=63.55, M_Fe=55.85)

   Moles Cu = 15.9 ÷ 63.55 = 0.2503 mol. Mole ratio Fe:Cu = 1:1 → moles Fe = 0.2503. Mass Fe = 0.2503 × 55.85 = 13.99 g
2. Mass of metal from ore (percent purity)
   
   An ore contains 12.0% Cu by mass. How much ore is required to obtain 24.0 g of pure copper (assuming 100% recovery)?
   Let mass of ore = x. 0.12x = 24.0 → x = 24.0 ÷ 0.12 = 200.0 g
3. Corrosion mass loss (conceptual)
   
   If 10.0 g of Fe forms Fe₂O₃ (rust), what mass of oxygen has combined with the iron? (M_Fe2O3=159.69)
   Moles Fe₂O₃ formed per 2 mol Fe. Moles Fe consumed = 10.0 ÷ 55.85 = 0.1790 mol → moles Fe₂O₃ = 0.1790 ÷ 2 = 0.08950 mol.

   Mass O in Fe₂O₃ = mass total − mass Fe = 0.08950 × 159.69 − 10.0 = (14.30 − 10.0) = 4.30 g O
4. Reduction of metal oxide
   
   2Fe₂O₃ + 3C → 4Fe + 3CO₂. If 160.0 g Fe₂O₃ is reduced, find mass of Fe produced. (M_Fe2O3=159.69, M_Fe=55.85)
   Moles Fe₂O₃ = 160.0 ÷ 159.69 = 1.0019 mol. From stoichiometry, 2 mol Fe₂O₃ → 4 mol Fe, so 1 mol Fe₂O₃ → 2 mol Fe. Moles Fe = 1.0019 × 2 = 2.0038 mol.

   Mass Fe = 2.0038 × 55.85 = 111.9 g
5. Electroplating — moles and mass (basic)
   
   If 0.0100 mol Ag⁺ is reduced and plated as Ag, what mass of silver metal is deposited? (M_Ag=107.87)
   Mass Ag = 0.0100 × 107.87 = 1.0787 g
6. Volume of hydrogen from acid-metal reaction
   
   2Al + 6HCl → 2AlCl₃ + 3H₂. If 5.40 g Al reacts, find volume of H₂ at STP produced.
   Moles Al = 5.40 ÷ 26.98 = 0.2002 mol. From equation, 2 mol Al → 3 mol H₂, so moles H₂ = 0.2002 × (3/2) = 0.3003 mol.

   Volume H₂ = 0.3003 × 22.4 = 6.726 L
7. Alloy composition by mass
   
   An alloy contains 70% Cu and 30% Zn by mass. How much Cu and Zn are present in 250 g of alloy?
   Mass Cu = 0.70 × 250 = 175.0 g. Mass Zn = 0.30 × 250 = 75.0 g
8. Mass of metal from oxide by reduction (mass–mass)
   
   If 88.8 g of PbO is reduced to Pb, what mass of lead is obtained? (PbO → Pb + 1/2O₂; M_PbO=223.20, M_Pb=207.2)
   Moles PbO = 88.8 ÷ 223.20 = 0.3980 mol → moles Pb = 0.3980 mol. Mass Pb = 0.3980 × 207.2 = 82.47 g
9. Percent purity from product
   
   10.0 g of an ore of iron oxide gives 6.31 g of pure iron after reduction. Calculate percent iron in ore.
   Percent Fe = (6.31 ÷ 10.0) × 100 = 63.1%
10. Limiting reagent — displacement
    
    If 5.00 g Zn is added to 20.0 g 0.1 M CuSO₄ solution (assume all Cu²⁺ reacts), which is limiting? Reaction: Zn + CuSO₄ → ZnSO₄ + Cu. (M_Zn=65.38, volume=0.0200 L)
    Moles CuSO₄ = 0.1 × 0.0200 = 0.00200 mol → gives 0.00200 mol Cu. Moles Zn available = 5.00 ÷ 65.38 = 0.07645 mol. CuSO₄ (Cu²⁺) is limiting (0.00200 << 0.07645).
11. Electrochemical mass deposition (intro)
    
    If one mole of electrons reduces 0.5 mol Cu²⁺ to Cu, how many coulombs are needed to deposit 1.00 g Cu? (Faraday constant ≈ 96500 C mol^-1, M_Cu=63.55)
    Moles Cu to deposit = 1.00 ÷ 63.55 = 0.01574 mol. Cu²⁺ requires 2 electrons per Cu → moles e^- = 0.01574 × 2 = 0.03148 mol e^-.

    Charge = 0.03148 × 96500 ≈ 3036 C (classroom-level calculation; show work on Faraday's law).
12. Gas volume from metal + acid
    
    2Mg + 2HCl → 2MgCl₂ + H₂. If 4.86 g Mg reacts, find mass and volume (STP) of H₂ produced.
    Moles Mg = 4.86 ÷ 24.305 = 0.2000 mol → moles H₂ = 0.1000 mol (from eqn 2:2:1 so 1 mol Mg gives 0.5 mol H₂).

    Mass H₂ = 0.1000 × 2.016 = 0.2016 g. Volume = 0.1000 × 22.4 = 2.24 L
13. Preparing metal salt — mass relations
    
    2Al + 6HCl → 2AlCl₃ + 3H₂O. If 5.40 g Al reacts with excess HCl, find mass AlCl₃ formed (M_AlCl3=133.34)
    Moles Al = 5.40 ÷ 26.98 = 0.2002 mol → moles AlCl₃ = 0.2002 mol (2:2). Mass = 0.2002 × 133.34 = 26.68 g
14. Mass balance in ore processing
    
    An ore contains 30% Fe by mass. From 500 g of ore, after processing, only 80% of iron is recovered. Find mass of iron obtained.
    Total Fe in ore = 0.30 × 500 = 150 g. Recovered = 0.80 × 150 = 120 g
15. Mass of metal from reacting compound
    
    If 9.30 g of ZnS is roasted to form ZnO and then reduced to Zn, what mass of Zn can be obtained? (ZnS → ZnO → Zn; M_ZnS=97.45, M_Zn=65.38)
    Moles ZnS = 9.30 ÷ 97.45 = 0.09545 mol → moles Zn = 0.09545 mol. Mass Zn = 0.09545 × 65.38 = 6.244 g
16. Mass percent in alloy
    
    An alloy of iron and chromium contains 12% Cr by mass. If 250 g alloy is used, find mass of Fe present.
    Mass Cr = 0.12 × 250 = 30.0 g → Mass Fe = 250 − 30 = 220 g
17. Stoichiometry — sulfide ore to metal
    
    2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, then Cu₂O → 2Cu + 1/2O₂. If 200 g Cu₂S is processed, find theoretical mass of Cu. (M_Cu2S=159.16, Cu atomic=63.55)
    Moles Cu₂S = 200 ÷ 159.16 = 1.2569 mol → each mole Cu₂S contains 2 Cu atoms → moles Cu = 1.2569 × 2 = 2.5138 mol.

    Mass Cu = 2.5138 × 63.55 = 159.7 g
18. Mass of oxide formed
    
    If 12.0 g of aluminium burns in excess oxygen to form Al₂O₃, find mass of oxide formed. (M_Al=26.98, M_Al2O3=101.96)
    Moles Al = 12.0 ÷ 26.98 = 0.4448 mol. Reaction: 4Al + 3O₂ → 2Al₂O₃; 4 mol Al → 2 mol Al₂O₃ so 1 mol Al → 0.5 mol Al₂O₃. Moles Al₂O₃ = 0.4448 × 0.5 = 0.2224 mol.

    Mass oxide = 0.2224 × 101.96 = 22.68 g
19. Percent yield — metal extraction
    
    Theoretical yield of copper from 100 g CuO is X g. If actual yield after reduction is 60.0 g, calculate percent yield. (M_CuO=79.545, M_Cu=63.55)
    Moles CuO = 100 ÷ 79.545 = 1.2579 mol → moles Cu = 1.2579 mol → theoretical mass Cu = 1.2579 × 63.55 = 79.92 g.

    Percent yield = (60.0 ÷ 79.92) × 100 = 75.06%
20. Calculation with ionic equation (simple)
    
    If 0.200 mol Zn reacts with excess H⁺ to form Zn²⁺ and H₂, what mass of H₂ is produced?
    Zn + 2H⁺ → Zn²⁺ + H₂. Moles H₂ = 0.200 ÷ 1 = 0.200 mol × (1/1)? (one mole Zn gives one mole H₂).

    Mass H₂ = 0.200 × 2.016 = 0.4032 g

Notes: Atomic/molar masses used are rounded to 3–4 significant figures for classroom clarity. Volumes at STP use 1 mol = 22.4 L. Problems follow NCERT/CBSE Class 10 style and are suitable for board exam practice.