Question

How many grams of aluminum are required to produce 3.5 miles Al2O3 in the presence of excess O2?

Answer 1

1) write the balanced equation:

4Al + 3O₂ ----> 2Al₂O₃

2) convert moles of Al₂O₃ to moles of Al using mole to mole ration

ration----> 4 mol Al = 2 mol Al₂O₃

3.5 mol Al₂O₃ (4 mol Al/ 2 mol Al₂O₃)= 7.0 mol Al

3) convert moles of Al to grams using its molar mass

molar mass of Al= 27.0 g/mol

7.0 mol Al (27.0 g/ 1 mol)= 189 grams