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

1. Neutralisation — basic balancing
   
   Balance: HCl + NaOH → NaCl + H₂O

   Solution: Balanced equation: HCl + NaOH → NaCl + H₂O (already balanced; 1:1 mole relation between HCl and NaOH).
2. Moles from mass — acid
   
   How many moles are there in 36.5 g of HCl? (M_HCl=36.46 g mol^-1)
   n = 36.5 ÷ 36.46 = 1.001 mol ≈ 1.00 mol
3. Mass of salt from neutralisation
   
   If 1.00 mol HCl completely neutralises 1.00 mol NaOH, find mass of NaCl formed. (M_NaCl=58.44 g mol^-1)
   Moles NaCl = 1.00 mol → mass = 1.00 × 58.44 = 58.44 g
4. Preparing a salt — mass required
   
   How many grams of NaOH are required to neutralise 73.0 g of HCl? (M_HCl=36.46, M_NaOH=40.00 g mol^-1)
   Moles HCl = 73.0 ÷ 36.46 = 2.002 mol.

   Moles NaOH required = 2.002 mol (1:1). Mass NaOH = 2.002 × 40.00 = 80.08 g
5. Molarity from mass and volume
   
   5.85 g of NaOH is dissolved to make 250.0 mL solution. Calculate molarity. (M_NaOH=40.00 g mol^-1)
   Moles NaOH = 5.85 ÷ 40.00 = 0.14625 mol.

   Volume = 0.2500 L → M = 0.14625 ÷ 0.2500 = 0.585 M
6. Volume of acid needed
   
   What volume of 0.200 M HCl is needed to neutralise 25.0 mL of 0.150 M NaOH? Reaction: HCl + NaOH → NaCl + H₂O
   Moles NaOH = 0.150 × 0.0250 = 0.00375 mol.

   Moles HCl needed = 0.00375 mol (1:1). Volume HCl = moles ÷ M = 0.00375 ÷ 0.200 = 0.01875 L = 18.75 mL
7. Mass of salt from reaction of metal carbonate
   
   CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. If 10.0 g CaCO₃ reacts with excess HCl, find mass of CaCl₂ produced (M_CaCO3=100.09, M_CaCl2=110.98 g mol^-1).
   Moles CaCO₃ = 10.0 ÷ 100.09 = 0.09991 mol.

   Moles CaCl₂ = 0.09991 mol (1:1). Mass CaCl₂ = 0.09991 × 110.98 = 11.09 g
8. Limiting reagent — acid and carbonate
   
   If 5.00 g Na₂CO₃ reacts with 10.0 g HCl, which is limiting? Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O (M_Na2CO3=105.99, M_HCl=36.46)
   Moles Na₂CO₃ = 5.00 ÷ 105.99 = 0.04717 mol. Requires 2×0.04717 = 0.09434 mol HCl.

   Moles HCl available = 10.0 ÷ 36.46 = 0.2743 mol. Since required < available, Na₂CO₃ is limiting.
9. Percent purity
   
   A sample of impure sodium carbonate weighing 15.0 g reacts with excess HCl to produce 6.24 g CO₂. If pure Na₂CO₃ would produce 4.405 g CO₂ per gram of Na₂CO₃, find percent purity.
   Moles CO₂ produced = 6.24 ÷ 44.01 = 0.1418 mol. Moles Na₂CO₃ consumed = 0.1418 mol (1:1).

   Mass pure Na₂CO₃ = 0.1418 × 105.99 = 15.02 g. Percent purity = (15.02 ÷ 15.0)×100 = 100.1% (≈100%) — indicates near pure sample (note: values chosen for classroom illustration).
10. Titration-style calculation
    
    50.0 mL of 0.100 M NaOH is titrated with 0.100 M HCl. What volume of HCl is needed for neutralisation?
    Moles NaOH = 0.0500 × 0.100 = 0.00500 mol. Moles HCl required = 0.00500 mol (1:1).

    Volume HCl = moles ÷ M = 0.00500 ÷ 0.100 = 0.0500 L = 50.0 mL
11. Mass of salt from reaction of metal with acid
    
    Mg + 2HCl → MgCl₂ + H₂. If 4.86 g Mg reacts fully, find mass of MgCl₂ produced (M_Mg=24.305, M_MgCl2=95.21).
    Moles Mg = 4.86 ÷ 24.305 = 0.2000 mol → Moles MgCl₂ = 0.2000 mol (1:1).

    Mass MgCl₂ = 0.2000 × 95.21 = 19.04 g
12. Concentration change after mixing
    
    10.0 mL of 1.00 M HCl is mixed with 40.0 mL of water. What is the resulting concentration of HCl?
    Moles HCl = 0.0100 × 1.00 = 0.0100 mol. Total volume = 0.0100 + 0.0400 = 0.0500 L.

    Molarity = 0.0100 ÷ 0.0500 = 0.200 M
13. Mass of salt from acid + base (different mole ratio)
    
    Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O. If 0.100 mol Al(OH)₃ reacts, find mass of AlCl₃ produced (M_AlCl3=133.34 g mol^-1).
    Moles AlCl₃ = 0.100 mol (1:1). Mass = 0.100 × 133.34 = 13.33 g
14. Mass of base from given acid
    
    How many grams of Ca(OH)₂ are required to neutralise 73.0 g of HCl? Reaction: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O (M_Ca(OH)2=74.09)
    Moles HCl = 73.0 ÷ 36.46 = 2.002 mol. Need 1 mol Ca(OH)₂ per 2 mol HCl → moles Ca(OH)₂ = 2.002 ÷ 2 = 1.001 mol.

    Mass Ca(OH)₂ = 1.001 × 74.09 = 74.16 g
15. Yield from reaction between acid and metal carbonate
    
    Theoretical mass of NaCl formed when 5.00 g Na₂CO₃ reacts with excess HCl. (M_NaCl=58.44)
    Moles Na₂CO₃ = 5.00 ÷ 105.99 = 0.04717 mol → produces 2 mol NaCl per mol Na₂CO₃ → moles NaCl = 0.09434 mol.

    Mass NaCl = 0.09434 × 58.44 = 5.513 g
16. Mass of gas evolved (CO2)
    
    If 20.0 g CaCO₃ reacts with excess HCl, find mass of CO₂ released.
    Moles CaCO₃ = 20.0 ÷ 100.09 = 0.1998 mol → Moles CO₂ = 0.1998 mol.

    Mass CO₂ = 0.1998 × 44.01 = 8.796 g
17. Mass of hydrated salt—conceptual
    
    Copper sulfate pentahydrate CuSO₄·5H₂O is formed when CuO reacts with H₂SO₄. If 10.0 g CuO reacts to give CuSO₄·5H₂O (M_CuO=79.545, M_CuSO4·5H2O=249.68), find mass of hydrated salt formed.
    Moles CuO = 10.0 ÷ 79.545 = 0.1257 mol → gives 0.1257 mol CuSO₄·5H₂O.

    Mass = 0.1257 × 249.68 = 31.39 g
18. Salts from acid + base mixture (mixed reactants)
    
    A mixture contains 0.100 mol HCl and 0.0500 mol HNO₃. If this mixture neutralises 0.150 mol NaOH completely, which acid is in excess or limiting?
    Total moles acid available (as H⁺) = 0.100 + 0.0500 = 0.150 mol. Since moles NaOH = 0.150 mol, they neutralise exactly → no excess; both consumed exactly.
19. Dilution calculation
    
    What volume of water must be added to 100.0 mL of 1.00 M HCl to make it 0.250 M?
    Moles HCl initially = 0.1000 × 1.00 = 0.1000 mol. Desired concentration 0.250 M → final volume = moles ÷ M = 0.1000 ÷ 0.250 = 0.400 L = 400.0 mL.

    Water to add = 400.0 − 100.0 = 300.0 mL
20. Percentage yield — salt preparation
    
    In preparation of NaCl by neutralising 5.85 g NaOH with excess HCl, the actual mass of NaCl obtained is 7.00 g. Calculate percent yield. (Use theoretical mass from reaction: NaOH + HCl → NaCl + H₂O; M_NaOH=40.00, M_NaCl=58.44)
    Moles NaOH = 5.85 ÷ 40.00 = 0.14625 mol → theoretical mass NaCl = 0.14625 × 58.44 = 8.548 g.

    Percent yield = (7.00 ÷ 8.548) × 100 = 81.89%

Notes: Molar masses used are rounded for classroom convenience. These problems follow the NCERT/CBSE style for Class 10 Chapter 2 and are suitable for board exam practice. Adjust significant figures per your marking scheme.