Aluminum and oxygen react according to the following equation: 4Al(s) + 3O 2(g) → 2Al2O3(s)
What mass of Al2O3, in grams, can be made by reacting 4.6 g Al with excess oxygen?
We’re being asked to calculate the mass of Al2O3, which will be formed by 4.6 g Al with excess O2. Since O2 is in excess, we can simply ignore it in our calculations. The flow for this problem will be like this:
Mass of Al (molar mass of Al) → Moles of Al (mole-to-mole comparison) → Moles of Al2O3 (molar mass of Al2O3) → Mass of Al2O3