Stoichiometry & Calculations — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Stoichiometry & Calculations MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Stoichiometry & Calculations · hard · numerical
For N₂ + 3H₂ → 2NH₃, 28 g of N₂ is mixed with 8 g of H₂. The maximum mass of NH₃ that can be formed is (N = 14, H = 1):
A. 68 g
B. 17 g
C. 45.3 g
D. 34 g ✓ Correct
Solution: Moles: N₂ = 28 ÷ 28 = 1.0, H₂ = 8 ÷ 2 = 4.0. One mol N₂ needs 3 mol H₂; here 1 mol N₂ needs 3 mol H₂ but 4 mol are available, so N₂ is limiting. NH₃ formed = 2 × 1 = 2 mol = 2 × 17 = 34 g. Wrongly treating H₂ as limiting (4 × 2/3 × 17) gives 45.3 g.
Q2 — Stoichiometry & Calculations · hard · numerical
For N₂ + 3H₂ → 2NH₃, 14 g of N₂ reacts with 4 g of H₂. After the reaction is complete, which reactant is left over and by what mass? (N = 14, H = 1)
A. H₂, 1 g remaining ✓ Correct
B. N₂, 1 g remaining
C. H₂, 2 g remaining
D. N₂, 2 g remaining
Solution: Moles: N₂ = 14 ÷ 28 = 0.5, H₂ = 4 ÷ 2 = 2.0. To consume 0.5 mol N₂ needs 3 × 0.5 = 1.5 mol H₂, and 2.0 mol are available, so N₂ is limiting and H₂ is in excess. H₂ used = 1.5 mol, leftover = 2.0 − 1.5 = 0.5 mol = 0.5 × 2 = 1 g of H₂. Not subtracting the H₂ used gives the wrong 2 g.
Q3 — Stoichiometry & Calculations · medium · numerical
For CH₄ + 2O₂ → CO₂ + 2H₂O, what volume of CO₂ measured at STP is produced by the complete combustion of 8.0 g of methane? (Molar volume at STP = 22.7 L mol⁻¹, C = 12, H = 1)
A. 22.7 L
B. 11.35 L ✓ Correct
C. 5.68 L
D. 11.2 L
Solution: Moles of CH₄ = 8.0 ÷ 16 = 0.5 mol. From the equation, 1 mol CH₄ gives 1 mol CO₂, so 0.5 mol CO₂ forms. Volume at STP = 0.5 × 22.7 = 11.35 L. Using the outdated 22.4 L mol⁻¹ gives the trap value 11.2 L.
Q4 — Stoichiometry & Calculations · hard · numerical
200 mL of 0.5 mol L⁻¹ HCl is mixed with 300 mL of 0.2 mol L⁻¹ HCl. Assuming volumes are additive, the molarity of the resulting solution is:
A. 0.35 mol L⁻¹
B. 0.32 mol L⁻¹ ✓ Correct
C. 0.16 mol L⁻¹
D. 0.70 mol L⁻¹
Solution: Moles of HCl = (0.200 × 0.5) + (0.300 × 0.2) = 0.10 + 0.06 = 0.16 mol. Total volume = 200 + 300 = 500 mL = 0.5 L. Molarity = 0.16 ÷ 0.5 = 0.32 mol L⁻¹. A plain average of the two molarities (0.5 + 0.2)/2 = 0.35 ignores the unequal volumes.
Q5 — Stoichiometry & Calculations · hard · numerical
Concentrated nitric acid is 63% HNO₃ by mass and has a density of 1.4 g mL⁻¹. Its molarity is (H = 1, N = 14, O = 16):
A. 14 mol L⁻¹ ✓ Correct
B. 10 mol L⁻¹
C. 22.2 mol L⁻¹
D. 9.9 mol L⁻¹
Solution: Take 1 L = 1000 mL of solution: mass = 1000 × 1.4 = 1400 g. Mass of HNO₃ = 63% of 1400 = 882 g. Molar mass HNO₃ = 1 + 14 + 48 = 63, so moles = 882 ÷ 63 = 14 mol in 1 L, giving 14 mol L⁻¹. Forgetting the density (using 1000 g instead of 1400 g) gives the wrong 10 mol L⁻¹.
Q6 — Stoichiometry & Calculations · hard · numerical
An aqueous solution of HCl is 36.5% HCl by mass. The molality of the solution is (H = 1, Cl = 35.5):
A. 10.0 mol kg⁻¹
B. 15.7 mol kg⁻¹ ✓ Correct
C. 27.4 mol kg⁻¹
D. 1.0 mol kg⁻¹
Solution: Take 100 g of solution: it has 36.5 g HCl and 100 − 36.5 = 63.5 g water. Moles of HCl = 36.5 ÷ 36.5 = 1.0 mol; mass of solvent = 63.5 g = 0.0635 kg. Molality = 1.0 ÷ 0.0635 = 15.7 mol kg⁻¹. Wrongly using 100 g (or the whole solution) as the solvent mass gives the trap 10.0 mol kg⁻¹.
Q7 — Stoichiometry & Calculations · hard · numerical
The mole fraction of ethanol (C₂H₅OH) in an aqueous solution is 0.25. If the density of the solution is 0.90 g mL⁻¹, its molarity is (C = 12, H = 1, O = 16):
A. 9.0 mol L⁻¹ ✓ Correct
B. 18.0 mol L⁻¹
C. 4.5 mol L⁻¹
D. 5.0 mol L⁻¹
Solution: Take 1 mol of solution: 0.25 mol ethanol and 0.75 mol water. Mass = 0.25 × 46 + 0.75 × 18 = 11.5 + 13.5 = 25 g. Volume = 25 ÷ 0.90 = 27.78 mL = 0.02778 L. Molarity = moles ethanol ÷ volume = 0.25 ÷ 0.02778 = 9.0 mol L⁻¹.