A thermite mixture contains 13.5 g of Al powder and 64 g of Fe2O3, and reacts according to the equation 2Al + Fe2O3 -> Al2O3 + 2Fe. Taking Al=27, Fe=56, O=16, the mass of iron formed when the limiting reagent is fully consumed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pb046

A thermite mixture contains $13.5 g$ of Al powder and $64 g$ of $Fe X_{2} O X_{3}$ , and reacts according to the equation $2 Al + Fe X_{2} O X_{3} Al X_{2} O X_{3} + 2 Fe$ . Taking $Al = 27$ , $Fe = 56$ , $O = 16$ , the mass of iron formed when the limiting reagent is fully consumed is

n (Al) = 13.5/27 = 0.50 mol

;

n (Fe X_{2} O X_{3}) = 64/160 = 0.40 mol

, which would need

0.80 mol

Al. So Al is the limiting reagent.

2 mol

Al give

2 mol

Fe, so

0.50 mol

= 0.50 \times 56 = 28.0 g

. (Using

Fe X_{2} O X_{3}

would wrongly give

44.8 g