Given the equation 4Al + 3O2 - -> 2Al2O3 if 325 grams of Al2O3 are to be formed, determine the mass of aluminum that must be reacted with excess oxygen. Question 4 options: 101.77g Al 26.9g Al 171.80g Al 6.38g Al

172 g Al

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let's gather all the information in one place.

M_r: 26.98 101.96

4Al + 3O₂ ⟶ 2Al₂O₃

m/g: 325

(a) Calculate the moles of Al₂O₃

n = 325 g Al₂O₃ * 1 mol Al₂O₃ / 39.10 g Al₂O₃

n = 3.188 mol Al₂O₃

(b) Calculate the moles of Al

The molar ratio is (4 mol Al/2 mol Al₂O₃)

n = 3.188 mol Al₂O₃ * (4 mol Al/2 mol Al₂O₃)

n = 6.375 mol Al

(c) Calculate the mass of Al

m = 6.375 mol Al * (26.98 g Al/1 mol Al)

m = 172 g Al

Note: The answer can have only three significant figures because that is all you gave for the mass of Al₂O₃.