Question

how many grams of oxygen gas (o2) are needed to completely react with 9.30 moles of aluminum? 4al + 3o2 → 2al2o3 9.30 mol al | 3 mol o2 | 32.0 g o2 | 4 mol al | 1 mol o2? g o2 mass o2 (g)

Explanation:

Step1: Calculate the number of moles of \(O_2\)

From the balanced chemical equation \(4Al + 3O_2
ightarrow2Al_2O_3\), the mole ratio of \(Al\) to \(O_2\) is \(4:3\).
Given \(n(Al)=9.30\space mol\), we use the mole - ratio to find \(n(O_2)\):
\(n(O_2)=9.30\space mol\space Al\times\frac{3\space mol\space O_2}{4\space mol\space Al}\)

$$n(O_2)=\frac{9.30\times3}{4}\space mol = 6.975\space mol$$

Step2: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M = 32.0\space g/mol\).
Using the formula \(m=n\times M\), where \(n = 6.975\space mol\) and \(M = 32.0\space g/mol\)
\(m(O_2)=6.975\space mol\times32.0\space g/mol\)

$$m(O_2)=6.975\times32.0\space g=223.2\space g$$

Answer:

\(223.2\space g\)