Stoichiometry & Calculations — NEET Chemistry MCQs with Solutions
Free NEET 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 · easy · numerical
When the equation C₃H₈ + O₂ → CO₂ + H₂O is balanced with the smallest whole-number coefficients, the coefficient of O₂ is:
A. 4
B. 5 ✓ Correct
C. 7
D. 3
Solution: Balance carbon first (3 CO₂), then hydrogen (4 H₂O): C₃H₈ + O₂ → 3CO₂ + 4H₂O. The right side now has 3×2 + 4×1 = 10 oxygen atoms, so 5 O₂ molecules are needed: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. The coefficient of O₂ is 5.
Q2 — Stoichiometry & Calculations · easy · numerical
For the combustion CH₄ + 2O₂ → CO₂ + 2H₂O, how many moles of methane are required to produce 22 g of CO₂? (C = 12, O = 16)
A. 0.25
B. 1.0
C. 2.0
D. 0.5 ✓ Correct
Solution: Molar mass of CO₂ = 12 + 2(16) = 44 g mol⁻¹, so 22 g CO₂ = 22 ÷ 44 = 0.5 mol. The equation shows 1 mol CH₄ gives 1 mol CO₂, so 0.5 mol CH₄ is required. Using 22 ÷ 88 gives the wrong 0.25 mol.
Q3 — Stoichiometry & Calculations · easy · numerical
4 g of NaOH is dissolved in enough water to make 250 mL of solution. The molarity of the solution is (Na = 23, O = 16, H = 1):
A. 0.8 mol L⁻¹
B. 0.4 mol L⁻¹ ✓ Correct
C. 0.04 mol L⁻¹
D. 0.1 mol L⁻¹
Solution: Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹, so moles = 4 ÷ 40 = 0.1 mol. Volume = 250 mL = 0.25 L. Molarity = moles ÷ volume in litres = 0.1 ÷ 0.25 = 0.4 mol L⁻¹. Reporting 0.1 forgets to divide by the volume in litres.
Q4 — Stoichiometry & Calculations · easy · numerical
5 g of urea is dissolved in 45 g of water. The mass percent (w/w) of urea in the solution is:
A. 9.09%
B. 11.1%
C. 10% ✓ Correct
D. 5%
Solution: Mass percent = (mass of solute ÷ mass of solution) × 100. Mass of solution = 5 + 45 = 50 g. So mass % = (5 ÷ 50) × 100 = 10%. Dividing by the solvent mass instead (5 ÷ 45 × 100 = 11.1%) is the common error.
Q5 — Stoichiometry & Calculations · medium · numerical
For CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, what mass of CaCO₃ reacts completely with 25 mL of 0.75 mol L⁻¹ HCl? (Ca = 40, C = 12, O = 16)
A. 0.47 g
B. 3.75 g
C. 0.94 g ✓ Correct
D. 1.88 g
Solution: Moles of HCl = 0.75 × 0.025 = 0.01875 mol. From the equation, CaCO₃ : HCl = 1 : 2, so moles CaCO₃ = 0.01875 ÷ 2 = 0.009375 mol. Molar mass of CaCO₃ = 40 + 12 + 48 = 100, so mass = 0.009375 × 100 = 0.94 g. Forgetting the 1 : 2 ratio (treating it as 1 : 1) gives the wrong 1.88 g.
Q6 — Stoichiometry & Calculations · medium · numerical
1.8 g of glucose (molar mass 180 g mol⁻¹) is dissolved in 18 g of water. The mole fraction of glucose in the solution is (H = 1, O = 16):
A. 0.0099 ✓ Correct
B. 0.01
C. 0.099
D. 0.0011
Solution: Moles of glucose = 1.8 ÷ 180 = 0.01. Moles of water = 18 ÷ 18 = 1.0. Mole fraction of glucose = n(glucose) ÷ (n(glucose) + n(water)) = 0.01 ÷ (0.01 + 1.0) = 0.01 ÷ 1.01 = 0.0099. Using 0.01 ÷ 1.0 (leaving glucose out of the denominator) gives the wrong 0.01.
Q7 — Stoichiometry & Calculations · medium · numerical
A 2 mol L⁻¹ solution of NaOH has a density of 1.10 g mL⁻¹. The molality of the solution is (Na = 23, O = 16, H = 1):
A. 1.96 mol kg⁻¹ ✓ Correct
B. 2.18 mol kg⁻¹
C. 2.00 mol kg⁻¹
D. 1.82 mol kg⁻¹
Solution: Take 1 L of solution: it contains 2 mol NaOH = 2 × 40 = 80 g. Mass of 1 L solution = 1000 mL × 1.10 g mL⁻¹ = 1100 g. Mass of water (solvent) = 1100 − 80 = 1020 g = 1.020 kg. Molality = moles solute ÷ kg solvent = 2 ÷ 1.020 = 1.96 mol kg⁻¹. Assuming molality equals molarity gives the wrong 2.00.