A + 2B rightarrow AB_{2}. 36.0 g of 'A' (Mol | vedclass

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(A) $1$. Calculate initial moles:$n_A = \frac{36.0 \ g}{60 \ g \ mol^{-1}} = 0.6 \ mol$$n_B = \frac{56.0 \ g}{80 \ g \ mol^{-1}} = 0.7 \ mol$$2$. Dete

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$C$ and $D$ only

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(A) $1$. Calculate initial moles:$n_A = \frac{36.0 \ g}{60 \ g \ mol^{-1}} = 0.6 \ mol$$n_B = \frac{56.0 \ g}{80 \ g \ mol^{-1}} = 0.7 \ mol$$2$. Determine limiting reagent:According to the reaction $A + 2B ightarrow AB_{2}$, $1 \ mol$ of $A$ requires $2 \ mol$ of $B$.For $0.6 \ mol$ of $A$, required $B = 0.6 	imes 2 = 1.2 \ mol$.Since we have only $0.7 \ mol$ of $B$,'$B$' is the limiting reagent.Statement $(A)$ is incorrect.$3$. Calculate product formed:$0.7 \ mol$ of $B$ will produce $\frac{0.7}{2} = 0.35 \ mol$ of $AB_{2}$.Molar mass of $AB_{2} = 60 + (2 	imes 80) = 220 \ g \ mol^{-1}$.Mass of $AB_{2} = 0.35 \ mol 	imes 220 \ g \ mol^{-1} = 77.0 \ g$.Statement $(B)$ and $(C)$ are correct.$4$. Calculate unreacted '$A$':$B$ is limiting, so $0.7 \ mol$ of $B$ reacts with $\frac{0.7}{2} = 0.35 \ mol$ of $A$.Remaining moles of $A = 0.6 - 0.35 = 0.25 \ mol$.Mass of remaining $A = 0.25 \ mol 	imes 60 \ g \ mol^{-1} = 15.0 \ g$.Statement $(D)$ is correct.Thus, statements $(B)$, $(C)$, and $(D)$ are correct. Based on the options provided, $(C)$ and $(D)$ are correct.

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