5 mol of aluminium chloride is mixed with 280 g of magnesium oxide, to give aluminium oxide. Find the amount of the excess reagent left after the reaction.
$2 A l C l_{3} + 3 M g O \to A l_{2} O_{3} + 3 M g C l_{2}$

44.055 g

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42.032 g

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54.056 g

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32.056 g

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Solution

The correct option is A 44.055 g
$2 A l C l_{3} + 3 M g O \to A l_{2} O_{3} + 3 M g C l_{2}$
Moles of $M g O = \frac{280}{40} = 7$ mol of MgO
2 mol of $A l C l_{3}$ reacts with 3 mol of MgO
5 mol will react with = 7.5 mol
So, MgO is the limiting reactant.
3 mol of MgO requires 2 mol of $A l C l_{3}$
7 mol of MgO will require = 4.67 mol
Excess reagent left = 5 – 4.67 = 0.33 mol
Mass of $A l C l_{3}$ left $= 0.33 \times 133.5 = 44.055$ g

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