Answer:
182 grams of oxygen gas must react to form 3.80 mol of aluminum oxide.
Explanation:
[tex]4Al + 3O_2\rightarrow 2Al_2O_3[/tex]
Moles of aluminum oxide = 3.80 mole
According to reaction, 2 moles of aluminum oxide is obtained from 3 moles of oxygen gas.
Then 3.80 moles of aluminum oxide will be obtained from:
[tex]\frac{3}{2}\times 3.80 mol=5.7 mol[/tex] of oxygen gas.
Mass of 3.80 moles of oxygen = 5.7 mol × 32 g/mol = 182.4 g ≈ 182 g
182 grams of oxygen gas must react to form 3.80 mol of aluminum oxide.
