Explanation

Number of moles of A is:

\[ n_A = \frac{36}{60} = 0.6\ \text{mol} \]

Number of moles of B is:

\[ n_B = \frac{56}{80} = 0.7\ \text{mol} \]

According to the reaction:

\[ A + 2B \rightarrow AB_2 \]

For 0.6 mol of A, required moles of B are:

\[ 2 \times 0.6 = 1.2\ \text{mol} \]

But only 0.7 mol of B is available, therefore B is the limiting reagent.

Moles of AB₂ formed are:

\[ \frac{0.7}{2} = 0.35\ \text{mol} \]

Molar mass of AB₂ is:

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

Mass of AB₂ formed:

\[ 0.35 \times 220 = 77.0\ \text{g} \]

Moles of A consumed:

\[ 0.35\ \text{mol} \]

Mass of A consumed:

\[ 0.35 \times 60 = 21.0\ \text{g} \]

Mass of A left unreacted:

\[ 36.0 - 21.0 = 15.0\ \text{g} \]

Hence, statements B and D are correct.

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