Question

\(A + 2B \longrightarrow AB_2\)
\(36.0\,\text{g}\) of \(A\) (Molar mass \(= 60\,\text{g mol}^{-1}\)) and \(56.0\,\text{g}\) of \(B\) (Molar mass \(= 80\,\text{g mol}^{-1}\)) are allowed to react. Which of the following statements are correct?
[A.] \(A\) is the limiting reagent.
[B.] \(77.0\,\text{g}\) of \(AB_2\) is formed.
[C.] Molar mass of \(AB_2\) is \(140\,\text{g mol}^{-1}\).
[D.] \(15.0\,\text{g}\) of \(A\) 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

Hint:

Always compare mole ratios with stoichiometric coefficients to identify the limiting reagent.

Answer 1

The Correct Option is C

Step 1: Calculate moles of reactants.
\[ n(A) = \frac{36.0}{60} = 0.6\ \text{mol} \] \[ n(B) = \frac{56.0}{80} = 0.7\ \text{mol} \]
Step 2: Identify the limiting reagent.
From the reaction, \[ A + 2B \rightarrow AB_2 \] \(0.6\) mol of \(A\) requires \(1.2\) mol of \(B\), but only \(0.7\) mol of \(B\) is available. Hence, \(B\) is the limiting reagent, and statement A is false.

Step 3: Calculate amount of product formed.
From stoichiometry, \[ 2\,\text{mol } B \rightarrow 1\,\text{mol } AB_2 \] \[ 0.7\,\text{mol } B \rightarrow 0.35\,\text{mol } AB_2 \] Molar mass of \(AB_2 = 60 + 2(80) = 220\,\text{g mol}^{-1}\) \[ \text{Mass of } AB_2 = 0.35 \times 220 = 77.0\,\text{g} \] Thus, statement B is correct, and statement C is false.

Step 4: Calculate unreacted \(A\).
Moles of \(A\) consumed: \[ 0.35\,\text{mol} \] Remaining moles of \(A\): \[ 0.6 - 0.35 = 0.25\,\text{mol} \] Mass of unreacted \(A\): \[ 0.25 \times 60 = 15.0\,\text{g} \] So, statement D is correct.

Final Answer: \[ \boxed{\text{B and D only}} \]