$13.5 g$ of $Al$ and $32.0 g$ of $Fe X_{2} O X_{3}$ ( $160 g/mol$ ) react as $2 Al + Fe X_{2} O X_{3} Al X_{2} O X_{3} + 2 Fe$ , but the reaction proceeds only to $90%$ of the extent possible for the limiting reagent. Taking $Al = 27$ , find the mass of $Al$ remaining unreacted, in grams, to two decimals.

Numerical answer type, answer in g, tolerance 0.02.

Answer: 3.78 g

Al

= 0.5 mol;

Fe X_{2} O X_{3}

= 0.2 mol, which needs 0.4 mol

Al

(available 0.5), so

Fe X_{2} O X_{3}

limits. Extent = 0.90 x 0.2 = 0.18 mol.

Al

used = 2 x 0.18 = 0.36 mol; left = 0.14 mol x 27 = 3.78 g.