In the reaction: $2 A l + C r_{2} O_{3} \to A l_{2} O_{3} + 2 C r$ , $4.98$ g of $A l$ reacted with $20.0$ g $C r_{2} O_{3}$ . How much grams of reactant remains at the completion of the reaction?

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Solution

Atomic weight of $A l$ and $C r = 27$ and $52$ respectively, Molecular weight of $C r_{2} O_{3} = 152$
Moles of $A l = \frac{4.98}{27 g A l} = 0.184 m o l$ $; \frac{0.184}{2} = 0.92$ mol of $C r_{2} O_{3}$
Moles of $C r_{2} O_{3} = \frac{20 g}{152 g C r_{2} O_{3}} = 0.131 m o l$
Since, $2$ mol $A l$ is required for $1$ mol of $C r_{2} O_{3}$ .
So, $A l$ is the limiting reagent and $C r_{2} O_{3}$ is in excess.
Moles of $C r_{2} O_{3}$ in excess $= (0.131 - 0.092) = 0.039 \approx 0.04 m o l$
Weight of $C r_{2} O_{3}$ in excess $= 0.04 \times 152 \approx 6 g C r_{2} O_{3}$ .
Thus, $6$ g of $C r_{2} O_{3}$ remains unreacted at the end of the reaction.

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