Question

what mass of al is needed to react with 16 g o₂?
4al + 3o₂ → 2al₂o₃
(16 g o₂) (1 mol o₂ / 32 g o₂) (4 mol al / 3 mol o₂) * (27 g al / 1 mol al) = ? g al
14 g al 18 g al 10 g al

Explanation:

Step1: Calculate moles of \( O_2 \)

Moles of \( O_2 = \frac{16\ g}{32\ g/mol} = 0.5\ mol \)

Step2: Relate moles of \( O_2 \) to \( Al \)

From reaction, \( 3\ mol\ O_2 \) reacts with \( 4\ mol\ Al \). So moles of \( Al = 0.5\ mol\ O_2 \times \frac{4\ mol\ Al}{3\ mol\ O_2} = \frac{2}{3}\ mol \)

Step3: Calculate mass of \( Al \)

Mass of \( Al = \frac{2}{3}\ mol \times 27\ g/mol = 18\ g \)

Answer:

18 g Al (corresponding to the middle option: 18 g Al)