⚗️ Chapter 15 · Stoichiometry (complete exercise including CRQs)

MCQs, short questions, constructed response & numericals · @everexams.com

A · MCQ (1–8) detailed solutions

1. Volume occupied by 15 moles NH₃ at RTP?

(a) 224 dm³ (b) 360 dm³ (c) 160 dm³ (d) 265 dm³

✅ Correct: (b) 360 dm³

Tip: At RTP, molar volume = 24 dm³/mol. V = 15×24 = 360 dm³.

2. Moles of CH₄ in 18 dm³ at RTP?

(a) 0.75 (b) 0.67 (c) 0.85 (d) 0.56

✅ (a) 0.75 mol

Moles = volume / 24 = 18/24 = 0.75 mol.

3. Molar conc. KOH: 2.0g in 100 cm³?

(a) 0.36 (b) 0.53 (c) 0.70 (d) 0.45 mol/dm³

✅ (a) 0.36 mol/dm³

Molar mass KOH=56; moles=2/56=0.0357; conc=0.0357/0.1 = 0.357 ≈0.36 M

4. Mass NaCl for 15 dm³ of 0.7 g/dm³?

(a) 21 g (b) 10.5 g (c) 31.5 g (d) 5.4 g

✅ (b) 10.5 g

Mass = conc (g/dm³) × volume = 0.7 × 15 = 10.5 g

5. Conc H₂SO₄ if 10 cm³ reacts with 20 cm³ 0.1 M NaOH?

(a) 0.01 (b) 0.001 (c) 0.1 (d) 0.15 M

✅ (c) 0.1 M

H₂SO₄ + 2NaOH → ; n(NaOH)=0.02×0.1=0.002 mol; n(H₂SO₄)=0.001 mol; conc=0.001/0.01=0.1M

6. Molar mass 178, empirical C₃H₅O₃? molecular formula?

(a) C₃H₅O₃ (b) C₆H₁₀O₆ (c) C₉H₁₅O₉ (d) C₈H₁₀O₈

✅ (b) C₆H₁₀O₆

Empirical mass = 3×12+5+48=89; n=178/89=2; mol = C₆H₁₀O₆

7. Natural gas burns, limiting reactant?

(a) both (b) neither (c) air (d) natural gas

✅ (c) Air (oxygen is usually limiting in open burning)

In air, O₂ is less than stoichiometric amount → limiting.

8. Actual 0.198g, theoretical 0.217g, % yield?

(a) 50% (b) 91.2% (c) 80% (d) 90%

✅ (b) 91.2%

(0.198/0.217)×100 = 91.24%

Quick key: 1-b,2-a,3-a,4-b,5-c,6-b,7-c,8-b

B · Short answer questions (key terms)

15.1 What is molar volume of a gas at RTP? 24 dm³/mol (or 24 L/mol) Volume occupied by 1 mole of any gas at room temperature and pressure (20°C, 1 atm).

15.2 How is molar volume useful? mole-volume conversion, gas calculations Allows calculation of gas volume from moles and vice versa without mass.

15.3 Relation between molar volume & Avogadro’s law? equal moles → equal volume at same T,P Avogadro: V ∝ n; molar volume = constant (RT/P) at fixed T,P.

15.4 Why two compounds show same empirical formula? same simplest ratio, different molecular mass E.g. C₂H₂ and C₆H₆ both have empirical CH.

C · Constructed Response Questions (detailed)

CRQ 1: A student reacts 50 g of zinc with excess nitric acid and obtains 80.2 g of Zn(NO₃)₂. Explain why the actual yield is less than theoretical. Calculate percentage yield and suggest two reasons for the difference.

Solution: Actual yield is always less due to incomplete reactions, side reactions, or loss during purification. Balanced eq: Zn + 2HNO₃ → Zn(NO₃)₂ + H₂. Molar mass Zn = 65.37 g/mol → moles Zn = 50/65.37 = 0.765 mol. Theoretical yield Zn(NO₃)₂ = 0.765 × (65.37+124) = 0.765 × 189.37 ≈ 144.9 g. % yield = (80.2/144.9)×100 = 55.35%. Reasons: (i) reaction not 100% complete, (ii) mechanical loss during filtration/transfer, (iii) side products.

CRQ 2: When 12.0 g sample of a drug contains 11.57 g active ingredient, calculate purity. Why is purity important in pharmaceuticals?

Solution: % purity = (mass of pure / total mass) × 100 = (11.57/12.0)×100 = 96.4%. Importance: ensures correct dosage, safety, efficacy, and compliance with regulatory standards.

CRQ 3: Explain the difference between actual yield, theoretical yield, and percentage yield. Use an example.

Solution: Theoretical yield is maximum product calculated from balanced equation (assuming 100% reaction). Actual yield is what is experimentally obtained. % yield = (actual/theoretical)×100. Example: In above zinc reaction, theoretical = 144.9 g, actual = 80.2 g → % = 55.35%.

CRQ 4: A 3.66 kg metal sample contains 2.45 kg gold. Find % purity. If impurities are 1.21 kg, what mass of pure gold would be in a 10 kg sample from the same ore?

Solution: % purity = (2.45/3.66)×100 = 66.94%. For 10 kg sample, pure gold = (66.94/100)×10 = 6.694 kg.

CRQ 5: When 12 g H₂ reacts with 64 g N₂ to form NH₃, identify limiting reactant and calculate mass of NH₃ formed theoretically.

Solution: 3H₂ + N₂ → 2NH₃. Moles H₂ = 12/2 = 6 mol; moles N₂ = 64/28 = 2.286 mol. For 2.286 mol N₂, need 3×2.286 = 6.858 mol H₂, but only 6 mol H₂ → H₂ limiting. From 6 mol H₂, moles NH₃ = (2/3)×6 = 4 mol → mass NH₃ = 4×17 = 68 g.

CRQ 6: A compound contains 40.0% C, 6.71% H, 53.3% O. 0.320 mol weighs 28.8 g. Determine molecular formula.

Solution: Molar mass = 28.8/0.320 = 90 g/mol. Empirical: per 100 g: C=40/12=3.333, H=6.71/1=6.71, O=53.3/16=3.331. Divide by smallest (3.33): C=1, H≈2.01, O=1 → empirical CH₂O (mass=30). n = 90/30 = 3 → molecular = C₃H₆O₃.

D · Numerical problems (solved)

15.1 4.0 g CH₄ at RTP: find moles, volume, molecules.
Molar mass CH₄ = 16 g/mol. Moles = 4/16 = 0.25 mol. Volume = 0.25 × 24 = 6.0 dm³. Molecules = 0.25 × 6.022×10²³ = 1.5055×10²³.

15.2 Mass NaCl in 150 cm³ of 0.4 mol/dm³ solution?
Volume = 150/1000 = 0.15 dm³. Moles = 0.4×0.15 = 0.06 mol. Mass = 0.06 × 58.5 = 3.51 g.

15.3 % sodium in NaHCO₃?
Molar mass = 23+1+12+48 = 84 g/mol. % Na = (23/84)×100 = 27.38%.

15.4 Iron sulphide: 1.926 g S, 2.333 g Fe. Empirical formula?
Moles Fe = 2.333/55.85 ≈ 0.04177; S = 1.926/32.06 ≈ 0.06007. Ratio Fe:S = 0.04177:0.06007 → divide by 0.04177 → Fe=1, S≈1.438 ≈ 1.44 → multiply by 2 → Fe₂S₃ (since 1.44×2=2.88 ≈3). Empirical = Fe₂S₃.

15.5 40.0% C,6.71%H,53.3%O; 0.320 mol weighs 28.8 g → molecular formula?
Molar mass = 28.8/0.32 = 90. Empirical CH₂O (mass30). n=90/30=3 → C₃H₆O₃.

15.6 12g H₂ + 64g N₂ → NH₃. Limiting reactant? (3H₂+N₂→2NH₃)
Moles H₂=6, N₂=2.286. Need 6.858 mol H₂ for all N₂ → H₂ limiting.

15.7 305 g AgNO₃ + excess MgCl₂ → 23.7 g Mg(NO₃)₂. % yield? 2AgNO₃ + MgCl₂ → Mg(NO₃)₂ + 2AgCl
Molar mass AgNO₃=170 → moles=305/170=1.794. 2 mol AgNO₃ → 1 mol Mg(NO₃)₂ → theoretical = 1.794/2 = 0.897 mol. Mg(NO₃)₂ molar mass≈148.3 → theoretical mass = 0.897×148.3≈133.0 g. % yield = (23.7/133.0)×100 = 17.82%.

15.8 Metal sample 3.66 kg, 2.45 kg gold. % purity = (2.45/3.66)×100 = 66.94%.

All numericals include essential steps & reasoning.

Student guidelines & memory anchors

✔️ Molar volume at RTP = 24 dm³/mol (like 24 cans of soda)
✔️ % yield = (actual/theoretical)×100 — actual is smaller
✔️ Limiting reactant: the one that gives least product
✔️ Empirical formula = simplest ratio, molecular = (empirical)×n
✔️ For CRQs: always define terms, show equation, then calculate, then reason.

✔️ Concentration (mol/dm³) = moles / volume(dm³)
✔️ For titration: use c₁v₁ / n₁ = c₂v₂ / n₂
✔️ Purity % = (pure/impure)×100
✔️ Constructed response: explain ‘why’ with scientific reasoning.