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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3mc199

$5.4 g$ of Al reacts with $10.65 g$ of $Cl X_{2}$ : $2 Al + 3 Cl X_{2} 2 AlCl X_{3}$ (Al = 27, Cl = 35.5). Select the correct statement(s).

A $13.35 g$ of $AlCl X_{3}$ is formed.
B $Cl X_{2}$ is the limiting reagent.
CAl is present in $100%$ excess over the amount needed to react with all the $Cl X_{2}$ .
D $1.35 g$ of Al remains unreacted.

Answer: A, B, C

n(Al) = 0.20 mol, n(Cl2) = 0.15 mol; Al needs 0.30 mol Cl2, so Cl2 is limiting. AlCl3 = 0.100 mol x 133.5 = 13.35 g. Al used = 0.100 mol, left = 0.100 mol = 2.7 g (not 1.35 g), which is 0.100/0.100 = 100% excess.

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