5.4 g of aluminium reacts with excess chlorine, 2Al + 3Cl2 -> 2AlCl3, and 20.0 g of AlCl3 is actually isolated (Al = 27, Cl = 35.5). The percentage yield is:

JEE Main levelSingle correct optionDifficulty 2 of 3mc042

$5.4 g$ of aluminium reacts with excess chlorine, $2 Al + 3 Cl X_{2} 2 AlCl X_{3}$ , and $20.0 g$ of $AlCl X_{3}$ is actually isolated (Al = 27, Cl = 35.5). The percentage yield is:

n (Al) = 5.4/27 = 0.20

mol, so theoretical

AlCl X_{3}

= 0.20 \times 133.5 = 26.70

g. Yield

= 20.0/26.70 \times 100 = 74.9%

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