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

1. Balancing a simple equation
   
   Balance: Fe + O₂ → Fe₂O₃

   Solution (quick steps): Fe appears as Fe₂ on product side → need 2 Fe atoms on reactant side. O atoms: O₂ gives 2 O; Fe₂O₃ has 3 O. Smallest whole-number coefficients: 4 Fe + 3 O₂ → 2 Fe₂O₃
2. Number of moles
   
   How many moles are present in 10.0 g of CaCO₃? (M_CaCO3 = 100.09 g mol^-1)
   n = mass / M = 10.0 ÷ 100.09 = 0.09991 mol
3. Mass of product from decomposition
   
   If 5.00 g of CaCO₃ decomposes: CaCO₃ → CaO + CO₂, find mass of CO₂ formed.
   1 mol CaCO₃ → 1 mol CO₂. M_CaCO3=100.09, M_CO2=44.01 g mol^-1.

   Moles CaCO₃ = 5.00 ÷ 100.09 = 0.04995 mol.

   Mass CO₂ = 0.04995 × 44.01 = 2.199 g (rounded to 3 d.p.)
4. Mass from a mass–mass calculation
   
   From the reaction 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu, how many grams of Cu are produced from 54.0 g of CuSO₄?
   M_CuSO4=159.61 g mol^-1, M_Cu=63.55 g mol^-1. Ratio Cu : CuSO₄ = 1 : 1 (from balanced equation).

   Moles CuSO₄ = 54.0 ÷ 159.61 = 0.3383 mol.

   Mass Cu = 0.3383 × 63.55 = 21.50 g
5. Gas volume at STP
   
   Decomposition of 44.0 g of CaCO₃ produces CO₂. Find the volume of CO₂ at STP (1 mol = 22.4 L).
   Moles CaCO₃ = 44.0 ÷ 100.09 = 0.4395 mol.

   Moles CO₂ = 0.4395 (1:1). Volume = 0.4395 × 22.4 = 9.85 L (3 s.f.)
6. Mass from gas reactant
   
   2.00 g of H₂ reacts with excess O₂ to form water. Find mass of H₂O produced. (M_H2=2.016, M_H2O=18.016)
   Moles H₂ = 2.00 ÷ 2.016 = 0.99206 mol.

   Reaction: 2H₂ → 2H₂O so moles H₂ = moles H₂O = 0.99206 mol.

   Mass H₂O = 0.99206 × 18.016 = 17.87 g
7. Number of particles
   
   How many molecules are there in 0.50 mol of H₂SO₄?
   Molecules = 0.50 × 6.022 × 10²³ = 3.011 × 10²³ molecules
8. Molarity from mass
   
   5.80 g of NaOH is dissolved to make 250.0 mL solution. Calculate molarity. (M_NaOH=39.998 g mol^-1)
   Moles NaOH = 5.80 ÷ 39.998 = 0.1450 mol.

   Volume = 0.2500 L → M = 0.1450 ÷ 0.2500 = 0.5800 M
9. Gas from metal–acid reaction
   
   Mg + 2HCl → MgCl₂ + H₂. If 3.00 g Mg reacts, find volume of H₂ at STP produced.
   Moles Mg = 3.00 ÷ 24.305 = 0.1234 mol. 1 mol Mg → 1 mol H₂.

   Volume H₂ = 0.1234 × 22.4 = 2.765 L
10. Mass of oxygen required
    
    CH₄ + 2O₂ → CO₂ + 2H₂O. How many grams of O₂ are required to burn 10.0 g of CH₄?
    Moles CH₄ = 10.0 ÷ 16.042 = 0.6235 mol.

    Moles O₂ required = 2 × 0.6235 = 1.2470 mol.

    Mass O₂ = 1.2470 × 32.00 = 39.90 g
11. Stoichiometry (organic combustion)
    
    Complete combustion: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. Find mass of CO₂ formed from 10.0 g glucose (M = 180.156 g mol^-1).
    Moles glucose = 10.0 ÷ 180.156 = 0.05553 mol.

    Moles CO₂ = 6 × 0.05553 = 0.3332 mol.

    Mass CO₂ = 0.3332 × 44.01 = 14.66 g
12. Large mass of gas — mass of water
    
    If 100.0 g of H₂ combines with oxygen to form water, what mass of H₂O is produced?
    Moles H₂ = 100.0 ÷ 2.016 = 49.603 mol.

    Moles H₂O = 49.603 mol (1:1 as 2H₂ → 2H₂O).

    Mass H₂O = 49.603 × 18.016 = 893.65 g
13. Limiting reagent
    
    10.0 g H₂ is mixed with 80.0 g O₂. For 2H₂ + O₂ → 2H₂O, determine the limiting reagent and mass of water formed.
    Moles H₂ = 10.0 ÷ 2.016 = 4.9603 mol. Moles O₂ = 80.0 ÷ 32.00 = 2.5000 mol.

    Required O₂ for 4.9603 mol H₂ = 4.9603 ÷ 2 = 2.4802 mol. Available O₂ = 2.5000 mol > required → H₂ is limiting.

    Moles H₂O produced = moles H₂ consumed = 4.9603 mol.

    Mass H₂O = 4.9603 × 18.016 = 89.37 g
14. Percentage yield
    
    If 10.0 g of CaCO₃ decomposes theoretically to give X g CO₂, but the actual mass obtained is 4.00 g, calculate percent yield.
    From earlier method: theoretical mass CO₂ = 10.0 ÷ 100.09 × 44.01 = 4.397 g.

    Percent yield = (actual ÷ theoretical) × 100 = (4.00 ÷ 4.397) × 100 = 90.97%
15. Mass → gas volume
    
    12.0 g carbon burns completely to CO₂. Find the volume of CO₂ at STP.
    Moles C = 12.0 ÷ 12.01 = 0.9992 mol → Moles CO₂=0.9992 mol.

    Volume = 0.9992 × 22.4 = 22.38 L
16. Metal–acid reaction (mass of H₂)
    
    Zn + 2HCl → ZnCl₂ + H₂. If 6.50 g Zn reacts fully, find mass of H₂ produced.
    Moles Zn = 6.50 ÷ 65.38 = 0.09945 mol. 1 mol Zn → 1 mol H₂.

    Mass H₂ = 0.09945 × 2.016 = 0.2004 g
17. Carbonate decomposition
    
    Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O. If 10.0 g Na₂CO₃ reacts, find mass of CO₂ produced (M_Na2CO3=105.99 g mol^-1).
    Moles Na₂CO₃ = 10.0 ÷ 105.99 = 0.09434 mol.

    Mass CO₂ = 0.09434 × 44.01 = 4.152 g
18. Combustion — ethane
    
    C₂H₆ + 3.5 O₂ → 2CO₂ + 3H₂O. Find mass of CO₂ from 2.00 g C₂H₆ (M = 30.068 g mol^-1).
    Moles C₂H₆ = 2.00 ÷ 30.068 = 0.06653 mol.

    Moles CO₂ = 2 × 0.06653 = 0.13306 mol.

    Mass CO₂ = 0.13306 × 44.01 = 5.855 g
19. Precipitation reaction (mass)
    
    If 0.20 mol AgNO₃ reacts with excess NaCl to form AgCl (AgNO₃ + NaCl → AgCl + NaNO₃), find mass of AgCl precipitated (M_AgCl=143.32 g mol^-1).
    Moles AgCl = 0.20 mol. Mass = 0.20 × 143.32 = 28.66 g
20. Formation of an ionic salt
    
    2K + Cl₂ → 2KCl. If 2.00 g of K reacts completely, find mass of KCl formed (M_K=39.10 g mol^-1, M_KCl=74.55 g mol^-1).
    Moles K = 2.00 ÷ 39.10 = 0.05115 mol. Each mole K gives 1 mole KCl (1:1 mole relation).

    Mass KCl = 0.05115 × 74.55 = 3.813 g

Notes: All molar masses used are rounded to 3–5 significant figures for classroom use. Volumes at STP are calculated using 1 mol gas = 22.4 L. These problems follow CBSE/NCERT style and difficulty for Class 10 board preparation.