JEE Main 2026 — Jan 22, Shift 2 — Chemistry solution

JEE Main2026Jan 22, Shift 2Chemistry

$A + 2 B ⟶ A B_{2}$

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 $A B_{2}$ is formed

(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

Moles: $n_{A} = \frac{36}{60} = 0.6$ mol, $n_{B} = \frac{56}{80} = 0.7$ mol.

Stoichiometry requires $A : B = 1 : 2$ . For 0.7 mol of B, only $0.35$ mol of A is needed — so B is the limiting reagent (A in excess). Statement (A) is wrong.

Moles of $A B_{2}$ formed $= \frac{n _{B}}{2} = 0.35$ mol. Molar mass of $A B_{2} = 60 + 2 \times 80 = 220$ g mol $^{- 1}$ , so (C) is wrong.

Mass of $A B_{2} = 0.35 \times 220 = 77$ g. Statement (B) is correct.

Unreacted A $= (0.6 - 0.35) \times 60 = 15$ g. Statement (D) is correct.

So B and D are correct.

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