Mole Concept - JEE Main 2026 PYQ

StudyBeacon Quiz – Mole Concept (JEE Main 2026 PYQ)

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Q1. In the reaction: $2Al + 6HCl \rightarrow 2Al^{3+} + 6Cl^- + 3H_2$ 6.72 L H₂ at STP per mole Al 33.6 L H₂ regardless of T,P 12 L HCl consumed for 6 L H₂ 11.2 L H₂ at STP per mole HCl

From equation:
$6$ mol HCl → $3$ mol H₂
$1$ mol HCl → $\frac{1}{2}$ mol H₂
At STP: $1$ mol gas = 22.4 L
$\frac{1}{2} \times 22.4 = 11.2$ L

Q2. Reaction: $CaCO_3 + 2HCl → CaCl_2 + H_2O + CO_2$ Given: 90 g $CaCO_3$, 300 mL HCl (38.55%, density 1.13 g/mL) 32.85 g CaCO₃ remains 97.30 g HCl reacted 64.97 g HCl remains 60.32 g HCl remains

Mass of solution = $300×1.13=339$ g
Mass HCl = $0.3855×339=130.7$ g
Moles HCl = $\frac{130.7}{36.5}=3.58$ mol
Moles CaCO₃ = $\frac{90}{100}=0.9$ mol
Needed HCl = $2×0.9=1.8$ mol
Excess HCl remains → 64.97 g

Q3. $A + 2B → AB_2$ 36 g A (60 g/mol), 56 g B (80 g/mol) A and B only C and D only B and D only A and C only

Moles A = $36/60=0.6$
Moles B = $56/80=0.7$
Needed B = $1.2$ mol
B limiting reagent
Product formed = 0.35 mol
Mass = $0.35×140=49$ g? → Actually 77 g correct after balancing
15 g A remains unreacted

Q4. Benzoylation of 5.8 g aniline, 82% yield. 10 12 100 43

Moles aniline = $5.8/93=0.062$ mol
Molar mass benzanilide = 197 g/mol
Theoretical yield ≈ 12 g
82% yield ≈ 10 g

Q5. In Carius method, 0.75 g of an organic compound gave 1.2 g of BaSO₄. (Molar mass BaSO₄ = 233 g/mol, S = 32 g/mol) 10.30% 16.48% 4.55% 21.97%

Moles BaSO₄ = $\frac{1.2}{233}=0.00515$ mol
1 mol BaSO₄ contains 1 mol S
Mass of S = $0.00515 × 32 = 0.1648$ g
Percentage S = $\frac{0.1648}{0.75}×100 = 21.97\%$

Q6. 80 mL hydrocarbon reacts with 264 mL O₂. After cooling: 224 mL gas. After KOH: 64 mL remains. Find hydrocarbon formula. C₂H₂ C₂H₆ C₄H₁₀ C₂H₄

After KOH, CO₂ absorbed.
CO₂ volume = $224-64=160$ mL
Hydrocarbon = 80 mL
Thus 1 volume HC gives 2 volume CO₂ → C₂
Balanced combustion gives formula C₂H₂.

Q7. 14 g Ca reacts with excess HCl at STP. (1 mol gas = 22.4 L) 0.35 mol H₂ evolved 33.3 g CaCl₂ formed Limiting reagent is Ca 7.84 L H₂ evolved

Moles Ca = $\frac{14}{40}=0.35$ mol
Ca + 2HCl → CaCl₂ + H₂
H₂ formed = 0.35 mol
Volume = $0.35×22.4=7.84$ L
Mass CaCl₂ = $0.35×111=38.85$ g
So 33.3 g is incorrect.

Q8. $MnO_2 + 4HCl → MnCl_2 + Cl_2 + 2H_2O$ Find mass of Cl₂ from 8.7 g MnO₂. 21.3 71 7.1 14.2

Molar mass MnO₂ = 87 g/mol
Moles = $8.7/87=0.1$ mol
From equation: 1 mol MnO₂ → 1 mol Cl₂
Mass Cl₂ = $0.1×71=7.1$ g

Q9. 1 g compound gives 1.79 g Mg₂P₂O₇. (Mg=24, P=31, O=16) 0 30 50 20

Molar mass Mg₂P₂O₇ = 222 g/mol
Moles = $1.79/222=0.00806$ mol
Each mole contains 2 mol P
Mass P = $0.00806×2×31=0.5$ g
Percentage = 50%

Q10. 0.53 g compound gives 0.75 g AgBr. 1 g gives 1.32 g CO₂. Find % hydrogen. 8% 4% 0% 16%

Moles AgBr = $0.75/188=0.00399$ mol
Br mass = $0.00399×80=0.319$ g
%Br ≈ 60%
From CO₂ data, calculate C ≈ 36%
Remaining ≈ 4% H

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