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(D) Given: $5 \ L$ of $2 \ M$ $AgNO_3$ solution.Reaction: $Al + 3Ag^{+} ightarrow Al^{3+} + 3Ag$.Total $Ag^{+}$ moles = $5 \ L 	imes 2 \ M = 10 \ m

Vedclass · Accepted Answer

If $81 \ g$ of $Al$ is oxidized,then $972 \ g$ of silver ions will remain unreduced in the solution at the end of the reaction.

Vedclass · Answer

(D) Given: $5 \ L$ of $2 \ M$ $AgNO_3$ solution.Reaction: $Al + 3Ag^{+} ightarrow Al^{3+} + 3Ag$.Total $Ag^{+}$ moles = $5 \ L 	imes 2 \ M = 10 \ mol$.$(A)$ $3 \ mol$ $Ag^{+}$ requires $1 \ mol$ $(27 \ g)$ $Al$. So,$10 \ mol$ $Ag^{+}$ requires $(10/3) 	imes 27 = 90 \ g$ $Al$. (Correct)$(B)$ $1 \ mol$ $(27 \ g)$ $Al$ reduces $3 \ mol$ $(3 	imes 108 = 324 \ g)$ $Ag^{+}$. Thus,$54 \ g$ $(2 \ mol)$ $Al$ reduces $6 \ mol$ $(6 	imes 108 = 648 \ g)$ $Ag^{+}$. (Correct)$(C)$ $1 \ mol$ $Al$ reduces $3 \ mol$ $(3 	imes 6.022 	imes 10^{23})$ $Ag^{+}$ ions. Thus,$5 \ mol$ $Al$ reduces $15 	imes 6.022 	imes 10^{23} = 90.33 	imes 10^{23}$ $Ag^{+}$ ions. (Correct)$(D)$ $81 \ g$ $(3 \ mol)$ $Al$ reduces $9 \ mol$ $(9 	imes 108 = 972 \ g)$ $Ag^{+}$. Total $Ag^{+}$ is $10 \ mol$ $(1080 \ g)$. Remaining $Ag^{+}$ = $1080 - 972 = 108 \ g$. The option states $972 \ g$ remains unreduced,which is incorrect. (Incorrect)

For the stoichiometric reaction occurring as shown in the figure,which option is incorrect?

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