A 20 mL sample of a sodium iodide solution yields 4.74 g of silver iodide precipitate when treated with an excess of…

Year	2026	2025	2024	Total
Questions	8	15	7	30

Q56 jee_main_2026_21_jan_morning Eudiometry

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

A. $C₂H₄$ $\mathrm{C}_2\mathrm{H}_4$
B. $C₄H₁₀$ $\mathrm{C}_4\mathrm{H}_{10}$
C. $C₂H₂$ $\mathrm{C}_2\mathrm{H}_2$
D. $C₂H₆$ $\mathrm{C}_2\mathrm{H}_6$

Solution

Related Formula

CₓHy(g) + (x + y4)O2(g) xCO2(g) + y2H₂O( )

Core Logic

Let the volume of hydrocarbon be $V = 80$ $V = 80$ mL. Initial volume of $O₂ = 264$ $O_2 = 264$ mL. At 273 K, $H₂O$ $H_2O$ is liquid, so its volume is neglected. Volume of $CO₂$ $CO_2$ formed = $80x$ $80x$ mL. Volume of $O₂$ $O_2$ used = $80(x + (y)/(4))$ $80\left(x + \frac{y}{4}\right)$ mL. Unreacted $O₂$ $O_2$ = $264 - 80(x + (y)/(4))$ $264 - 80\left(x + \frac{y}{4}\right)$ mL.

Total residual volume = $VCO₂ + Vunreacted O₂ = 224$ $V_{CO_2} + V_{unreacted \ O_2} = 224$ mL.

80x + 264 - 80(x + (y)/(4)) = 224

264 - (80y)/(4) = 224

40 = 20y y = 2

After treatment with KOH, $CO₂$ $CO_2$ is absorbed. The remaining volume is unreacted $O₂$ $O_2$, which is 64 mL.

264 - 80(x + (y)/(4)) = 64

Substitute $y = 2$ $y = 2$:

264 - 80(x + (1)/(2)) = 64

264 - 80x - 40 = 64

$224 - 80x = 64$ $224 - 80x = 64$

80x = 160 x = 2

The hydrocarbon is $C₂H₂$ $\mathrm{C}_2\mathrm{H}_2$.

Pattern Recognition

Volume decrease by KOH indicates the volume of $CO₂$ $CO_2$ produced. $VCO₂ = 224 - 64 = 160$ $V_{CO_2} = 224 - 64 = 160$ mL. $VHC = 80$ $V_{HC} = 80$ mL. So $x = (160)/(80) = 2$ $x = \frac{160}{80} = 2$. Total volume reduction = 264 - 64 = 200 mL ( $O₂$ $O_2$ consumed). $O₂$ $O_2$ consumed = $80(x + y/4) = 200 2 + y/4 = 2.5 y/4 = 0.5 y = 2$ .

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q57 jee_main_2026_21_jan_morning Stoichiometry and Limiting Reagent

14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect? [Given : Molar mass in

g mol⁻¹

of Ca–40, Cl–35.5, H–1]

A. $0.35 mol of H₂ gas is evolved.$
B. $7.84 ~L of H₂ gas is evolved.$
C. $33.3 g of CaCl₂ is produced.$
D. $The limiting reagent is calcium metal.$

Solution

Related Formula

Ca(s) + 2HCl(g) CaCl2(s) + H2(g)

Core Logic

Number of moles of Calcium ( $nCa$ $n_{Ca}$) = $Given massMolar mass = (14.0)/(40) = 0.35$ mol. Since HCl is in excess, Calcium is the limiting reagent. From stoichiometry, 1 mole of Ca produces 1 mole of $H₂$ $H_2$ gas and 1 mole of $CaCl₂$ $CaCl_2$.

Moles of $H₂$ $H_2$ evolved = 0.35 mol. Volume of $H₂$ $H_2$ at STP (1 atm, 273 K) = $0.35 \times 22.4$ $0.35 \times 22.4$ L = $7.84$ $7.84$ L. Mass of $CaCl₂$ $CaCl_2$ produced = $0.35 \times Molar mass of CaCl₂$ = $0.35 \times (40 + 71) = 0.35 \times 111 = 38.85$ g.

Option (3) states 33.3 g of $CaCl₂$ $CaCl_2$ is produced, which is incorrect.

Step 1: Final Conclusion

Statement (3) is incorrect.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q55 jee_main_2026_21_jan_evening Percentage Composition and Stoichiometry

By usual analysis, 1.00g of compound (X) gave 1.79g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer) (Given, molar mass in

g mol⁻¹: O = 16, Mg = 24, P = 31

)

A. $(1) 50$ $(1) \ 50$
B. $(2) 30$ $(2) \ 30$
C. $(3) 20$ $(3) \ 20$
D. $(4) 40$ $(4) \ 40$

Solution

Related Formula

% of P = Moles of Mg₂P₂O₇ \times 2 \times 31Mass of compound \times 100

Core Logic

Molar mass of $Mg₂P₂O₇ = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 g/mol$ .

Percentage of P = (((1.79)/(222) \times 2 \times 31))/(1) \times 100 = 49.99% \approx 50%

Step 1: Final Calculation

Rounding to the nearest integer gives 50.

Pattern Recognition

Sees: Gravimetric analysis involving magnesium pyrophosphate. Trap: Forgetting the factor of 2 for phosphorus atoms in $Mg₂P₂O₇$ $\text{Mg}_2\text{P}_2\text{O}_7$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q56 jee_main_2026_22_january_morning Stoichiometry and Molar Volume

In the reaction,

2Al(s) + 6HCl(aq) arrow 2Al³⁺(aq) + 6Cl⁻(aq) + 3H₂(g)

A. $11.2 L H₂(g) at STP is produced for every mole of HCl consumed .$
B. $67.2 L H₂(g) at STP is produced for every mole of Al that reacts.$
C. $12 L HCl(aq) is consumed for every 6L H₂(g) produced.$
D. $33.6 L H₂(g) is produced regardless of temperature and pressure for every mole of Al that reacts.$

Solution

Related Formula

Volume of gas at STP = Moles \times 22.4 L

Core Logic

From the balanced chemical equation:

2Al(s) + 6HCl(aq) arrow 2Al³⁺(aq) + 6Cl⁻(aq) + 3H₂(g)

6 moles of HCl produce 3 moles of $H₂$ $H_2$. Therefore, 1 mole of HCl produces $(3)/(6) = 0.5$ $\frac{3}{6} = 0.5$ moles of $H₂$ $H_2$.

Volume of $H₂$ $H_2$ produced at STP for 1 mole of HCl:

V = 0.5 \times 22.4 L = 11.2 L

Step 1: Check other options

Option (2): 2 moles of Al produce 3 moles of $H₂$ $H_2$. 1 mole Al produces 1.5 moles $H₂ = 1.5 \times 22.4 = 33.6 L$ $H_2 = 1.5 \times 22.4 = 33.6\text{ L}$ (Incorrect). Option (4) is incorrect because volume depends on temperature and pressure.

Pattern Recognition

Standard stoichiometry. Directly map the mole ratio from the balanced equation to molar volume at standard conditions.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q53 jee_main_2026_22_january_evening Limiting Reagent and Stoichiometry

A + 2B arrow AB₂

36.0 g

of 'A' (Molar mass:

60 g mol⁻¹

) and

56.0 g

of 'B' (Molar mass:

80 g mol⁻¹

) are allowed to react. Which of the following statements are correct? (A) 'A' is the limiting reagent (B)

77.0 g

of

AB₂

is formed (C) Molar mass of

AB₂

is

140 g mol⁻¹

(D)

15.0 g

of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below:

A. C and D only
B. A and C only
C. B and D only
D. A and B only

Solution

Related Formula

Moles (n) = Given MassMolar Mass

Molar Mass of AB₂ = MA + 2 MB

Core Logic

Step 1: Calculate initial moles:

nA = (36)/(60) = 0.6 mol

nB = (56)/(80) = 0.7 mol

Step 2: Identify Limiting Reagent (LR):

Ratio for A = (0.6)/(1) = 0.6, Ratio for B = (0.7)/(2) = 0.35

Since ratio of B is smaller, B is the limiting reagent.

Step 3: Evaluate product formed and remaining reactant:

Molar mass of AB₂ = 60 + 2(80) = 220 g mol⁻¹

Moles of AB₂ formed = 0.35 mol

Mass of AB₂ formed = 0.35 \times 220 = 77.0 g (Statement B is correct)

Moles of A reacted = 0.35 mol

Moles of A remaining = 0.6 - 0.35 = 0.25 mol

Mass of A remaining = 0.25 \times 60 = 15.0 g (Statement D is correct)

Pattern Recognition

Sees: Initial masses of reactants in stoichiometric equation. Shortcut: Compare $n/coefficient$ $n/\text{coefficient}$ to find LR, then compute formed product mass and unreacted mass directly.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

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