Question

4al + 3o2 → 2al2o3 how many grams of aluminum(al) are required to produce 3. 5 moles al2o3 in the presence of excess o2?.

Answer 1

According to the balanced reaction, one needs 4 mol Al for every 2 mol of Al₂O₃. That is, we need twice as much Al to produce a given amount of Al₂O₃. So we need 7.0 mol Al.

Convert this to a mass. The molar mass of Al is about 26.98 g/mol, so

(7.0 mol) (26.98 g/mol) ≈ 188.86 g ≈ 190 g