Answer:
52.8 g of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4Al + 3O2 —> 2Al2O3
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3
Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2.
Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.
Thus, 1.65 moles of O2 is needed for the reaction.
Finally, we shall determine the mass of O2 needed as shown below:
Mole of O2 = 1.65 moles
Molar mass of O2 = 2 × 16= 32 g/mol
Mass of O2 =?
Mole = mass/Molar mass
1.65 = mass of O2 /32
Cross multiply
Mass of O2 = 1.65 × 32
Mass of O2 = 52.8 g
Therefore, 52.8 g of O2 is needed for the reaction.
