Question

Consider the reaction $\text{2Al + 3I}\to 2\text{Al}{\text{I}}_{\text{3}}$
Identity the Limiting reagent (theoretical yield) of the product if one starts with:-
(I)12 mol of $Al$ + 2.4 mol $I$${}_2$
(ii) 1.2 $Al$ +2.4 g $I_2$
(III) How many grams of $Al$ are left mol?

Answer 1

$2Al+3I_2⟶2Al{I}_3$

$E^{max}=\frac{0-n_{input}}{-\upsilon \left(coefficient\right)}$

$\left(I\right)$ $2Al+3I_2⟶2Al{I}_3$

$12$ $mol$ $2.4$ $mol$

$10.4$ $mol$ $1.6$ $mol$

$LR$ is that compound which have minimum $E^{max}$ (Maximum extent of reaction)

For $Al,E^{max}=\frac{0-12}{-2}=6$

For $I_2,E^{max}=\frac{0-2.4}{-3}=0.8$

$⟹I_2$ is limiting.

$\left(II\right)$ $1.2$ $mole$ $Al+2.4$ $mole$ $I_2$

For $Al,E^{max}=\frac{0-1.2}{-2}=0.6$

For $I,E^{max}=\frac{0-2.4}{-3}=0.8$

$⟹Al$ is limiting.

$\left(III\right)$ $2Al+3I_2⟶2Al{I}_3⟶$ For part $II$

$1.2$ $2.4$

$0.6$ $1.2$

Since, $Al$ was limiting, no amount of $Al$ is left

For part $I$

$2Al+3I_2⟶2Al{I}_3$

$12$ $mol$ $2.4$ $mol$

$10.4$ $mol$ $1.6$ $mol$

Mass of $Al$ left $=10.4\times 27=280.8$ $g$