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JEE Chemistry 2026 Question with Solution

A+2BAB2A + 2B \longrightarrow AB_2

36.0g36.0 \, g of AA (Molar mass =60g mol1= 60 \, g \ mol^{-1}) and 56.0g56.0 \, g of BB (Molar mass =80g mol1= 80 \, g \ mol^{-1}) are allowed to react. Which of the following statements are correct?

[A.] AA is the limiting reagent. [B.] 77.0g77.0 \, g of AB2AB_2 is formed. [C.] Molar mass of AB2AB_2 is 140g mol1140 \, g \ mol^{-1}. [D.] 15.0g15.0 \, g of AA is left unreacted after completion of reaction.

Choose the correct answer from the options given below:

  • A

    A and B only

  • B

    A and C only

  • C

    B and D only

  • D

    C and D only

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Reaction is A+2BAB2A + 2B \rightarrow AB_2. Mass of AA is 36.0g36.0 \, g with molar mass 60g mol160 \, g \ mol^{-1}. Mass of BB is 56.0g56.0 \, g with molar mass 80g mol180 \, g \ mol^{-1}.

Find: Which statements among A, B, C, and D are correct.

Step 1: Calculate moles of reactants

n(A)=36.060=0.6 moln(A) = \frac{36.0}{60} = 0.6 \ \text{mol} n(B)=56.080=0.7 moln(B) = \frac{56.0}{80} = 0.7 \ \text{mol}

Step 2: Identify the limiting reagent From

A+2BAB2A + 2B \rightarrow AB_2

0.6mol0.6 \, mol of AA requires

2×0.6=1.2 mol of B2 \times 0.6 = 1.2 \ \text{mol of } B

but only 0.7mol0.7 \, mol of BB is available. Therefore, BB is the limiting reagent, so statement A is false.

Step 3: Calculate amount of product formed From stoichiometry,

2 mol B1 mol AB22 \ \text{mol } B \rightarrow 1 \ \text{mol } AB_2

so

0.7 mol B0.35 mol AB20.7 \ \text{mol } B \rightarrow 0.35 \ \text{mol } AB_2

Molar mass of AB2AB_2 is

60+2(80)=220 g mol160 + 2(80) = 220 \ \text{g mol}^{-1}

Therefore, mass of AB2AB_2 formed is

0.35×220=77.0 g0.35 \times 220 = 77.0 \ \text{g}

So statement B is correct, and statement C is false because the molar mass is 220g mol1220 \, g \ mol^{-1}, not 140g mol1140 \, g \ mol^{-1}.

Step 4: Calculate unreacted AA Moles of AA consumed are

0.35 mol0.35 \ \text{mol}

Remaining moles of AA are

0.60.35=0.25 mol0.6 - 0.35 = 0.25 \ \text{mol}

Mass of unreacted AA is

0.25×60=15.0 g0.25 \times 60 = 15.0 \ \text{g}

So statement D is correct.

Conclude: The correct statements are B and D only. Therefore, the correct option is C.

Statement-wise Verification

Given: A+2BAB2A + 2B \rightarrow AB_2

Find: Check each statement individually.

  1. Statement A: "AA is the limiting reagent."
  • Available moles: n(A)=0.6n(A)=0.6 and n(B)=0.7n(B)=0.7.
  • Required ratio is 1:21:2.
  • For 0.6mol0.6 \, mol of AA, required BB is 1.2mol1.2 \, mol.
  • Since only 0.7mol0.7 \, mol of BB is available, BB limits the reaction.
  • Hence statement A is false.
  1. Statement B: "77.0g77.0 \, g of AB2AB_2 is formed."
  • From 0.7mol0.7 \, mol of limiting reagent BB,
moles of AB2=0.72=0.35\text{moles of } AB_2 = \frac{0.7}{2} = 0.35
  • Molar mass of AB2AB_2 is
60+2(80)=220 g mol160 + 2(80) = 220 \ \text{g mol}^{-1}
  • Product mass is
0.35×220=77.0 g0.35 \times 220 = 77.0 \ \text{g}
  • Hence statement B is true.
  1. Statement C: "Molar mass of AB2AB_2 is 140g mol1140 \, g \ mol^{-1}."
  • Actual molar mass is
60+160=220 g mol160 + 160 = 220 \ \text{g mol}^{-1}
  • Hence statement C is false.
  1. Statement D: "15.0g15.0 \, g of AA is left unreacted after completion of reaction."
  • Moles of AA consumed with 0.7mol0.7 \, mol of BB are
0.72=0.35 mol\frac{0.7}{2} = 0.35 \ \text{mol}
  • Remaining AA is
0.60.35=0.25 mol0.6 - 0.35 = 0.25 \ \text{mol}
  • Remaining mass is
0.25×60=15.0 g0.25 \times 60 = 15.0 \ \text{g}
  • Hence statement D is true.

Conclude: Only statements B and D are correct, so the correct option is C.

Common mistakes

  • Assuming the reactant with smaller mass is the limiting reagent. This is wrong because limiting reagent depends on moles and stoichiometric coefficients, not directly on mass. First convert each mass to moles and then compare using the balanced equation.

  • Using a 1:11:1 mole ratio between AA and BB. This is wrong because the reaction is A+2BAB2A + 2B \rightarrow AB_2, so 22 moles of BB react with 11 mole of AA. Always use coefficients from the balanced equation.

  • Calculating molar mass of AB2AB_2 as 140g mol1140 \, g \ mol^{-1} by adding incorrectly. This is wrong because AB2AB_2 contains one AA and two BB atoms, so its molar mass is 60+2×80=220g mol160 + 2 \times 80 = 220 \, g \ mol^{-1}.

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