If 75 grams of oxygen react, how many grams of aluminum are required?

84.24 g

Explanation:

Given dа ta:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of oxygen:

Number of moles = mass / molar mass

Number of moles = 75 g / 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

O₂ : Al

3 : 4

2.34 : 4/3*2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles * molar mass

Mass = 3.12 mol * 27 g/mol

Mass = 84.24 g