What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed as shown in the following reaction? (NO CALCULATOR) Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s) (A) 13.5 g (B) 27.0 g (C) 40.0 g (D) 67.0 g

13.5

Explanation:

Molar mass of Fe2O3 = 2*56 + 16*3 = 112 + 48 =160 approximate value

Number of moles of Fe2O3 = mass/molar mass = 40/160 = 0.25 mol

The reaction is balanced. Fine

1 mol of Fe2O3> 2 mol of Al

0.25 mol> 0.25*2 = 0.5 mol of Al

Mass of Al required = Number of moles * Molar mass

= 0.5 * 27 = 13.5 g

the answer is symmetrical : )