Question

how many moles of o₂ are needed to completely react with 54.0 g of al? 4al + 3o₂ → 2al₂o₃

Explanation:

Step1: Calculate moles of Al

Molar mass of Al is \(26.98\space g/mol\).
Moles of \(Al=\frac{54.0\space g}{26.98\space g/mol}\approx 2.00\space mol\)

Step2: Use mole ratio from balanced equation

From \(4Al + 3O_{2}\to2Al_{2}O_{3}\), mole ratio of \(O_{2}\) to \(Al\) is \(\frac{3}{4}\).
Moles of \(O_{2}=2.00\space mol\space Al\times\frac{3\space mol\space O_{2}}{4\space mol\space Al}= 1.50\space mol\space O_{2}\)

Answer:

\(1.50\space mol\space O_{2}\)