3 free questions from this chapter (the full set has 30). Marking: +4 for correct · −1 for incorrect · 0 for unattempted.
Question 1Level 2 · ExamSingle correct
Heavy water is D2O where D is deuterium (12H). Taking the oxygen as 816O only, D = 2 u, O = 16 u and NA=6.022×1023 mol−1, the total number of neutrons present in 2.0 g of D2O is
A4.82×1023
B6.02×1023
C5.35×1023
D6.69×1023
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Answer: B
Key idea: both the molar mass and the neutron count change when H is replaced by D. Molar mass of D2O = 2(2) + 16 = 20 g/mol, so 2.0 g = 0.10 mol. Neutrons per molecule: each D has 1 neutron and 16O has 8, total 10. Neutrons = 0.10×10×6.022×1023=6.02×1023. Trap 1: forgetting the neutron in each D gives 8 per molecule and 4.82×1023. Trap 2: using M = 18 (ordinary water) gives 5.35×1023 (8 neutrons) or 6.69×1023 (10 neutrons). Exam pattern: JEE Main – counting electrons/protons/neutrons in a given mass of a molecule or ionQuestion 2Level 3 · AdvancedSingle correct
12.4 g of white phosphorus (P4) is treated with 56.8 g of Cl2. The reaction P4+6Cl2→4PCl3 goes to completion first; any chlorine left over then converts PCl3 into PCl5 by PCl3+Cl2→PCl5 until one of them is exhausted. Using P = 31, Cl = 35.5, the mass of PCl5 in the final mixture is
A20.9 g
B66.7 g
C41.7 g
D83.4 g
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Answer: C
Key idea: the limiting reagent must be decided step by step, not from an overall equation. n(P4)=12.4/124=0.10 mol, n(Cl2)=56.8/71=0.80 mol. Step 1: 0.10 mol P4 uses 0.60 mol Cl2 and gives 0.40 mol PCl3; 0.20 mol Cl2 remains. Step 2: now Cl2 (0.20) is limiting against PCl3 (0.40), giving 0.20 mol PCl5 and leaving 0.20 mol PCl3. Mass of PCl5 = 0.20 x 208.5 = 41.7 g. Trap: writing the overall P4+10Cl2→4PCl5 makes Cl2 limiting and gives 0.32 mol = 66.7 g; assuming all 0.40 mol PCl3 is converted gives 83.4 g. Exam pattern: JEE Main – limiting reagent with consecutive reactionsQuestion 3Level 3 · AdvancedNumerical answer
125 g of CaCO3 is heated; decomposition CaCO3→CaO+CO2 occurs with 80% yield. All the CaO obtained is heated with 28.8 g of coke (pure carbon): CaO+3C→CaC2+CO, this step having a 62.5% yield based on the limiting reagent. The CaC2 is treated with excess water: CaC2+2H2O→Ca(OH)2+C2H2 (100% yield). Using Ca = 40, C = 12, O = 16, H = 1, find the mass of C2H2 formed, in grams.
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Answer: 13
Key idea: yields multiply only along the path of the limiting reagent, and the limiting reagent can switch between steps. Step 1: n(CaCO3)=1.25 mol, so CaO = 0.80 x 1.25 = 1.00 mol. Step 2: carbon = 28.8/12 = 2.40 mol, which can react with only 2.40/3 = 0.80 mol CaO, so carbon (not CaO) is limiting; theoretical CaC2 = 0.80 mol; actual = 0.625 x 0.80 = 0.50 mol. Step 3: C2H2 = 0.50 mol = 0.50 x 26 = 13 g. Trap: carrying the CaO amount forward (1.00 x 0.625 = 0.625 mol) gives 16.25 g. Exam pattern: JEE Main – overall yield of consecutive reactions with a limiting reagent