HELP THIS IS URGENT

Question

4 Al(s) + 3 O2(g) → 2 Al2O3(s)

What is the minimum number of grams of aluminum metal required to produce 4.00 mole of aluminum oxide?

Answer 1

We can use dimensional analysis to take the given quantity and convert it to the quantity of interest:

In words: We begin with 4.00 moles of aluminum oxide, and there are 4 moles of aluminum per 2 moles of aluminum oxide in the balanced reaction. Finally, each mole of aluminum weighs 26.9815 grams.

With symbols: 4.00 mol Al₂O₃ * 4 mol Al/2 mol Al₂O₃ * 26.9815 g Al / 1 mol Al

Cancel units: 4.00 mol Al₂O₃ * 4 mol Al/2 mol Al₂O₃ * 26.9815 g Al / 1 mol Al = 215.852 g Al

With 3 sig figs: 215 g Al