Question

4A1 + 302 →

2Al2O3. How many grams of oxygen (O2)
are needed to produce 95 grams of
aluminum oxide (Al2O3)?

Answer 1

Answer: 44.706g Al2O3

Step-by-step explanation:

4A1 + 302 → 2Al2O3

3×(16×2) 2{(27×2)+(16×3)}

= 96 = 204

So

To produce 204g of Al2O3,We need . 96g of O2

To produce 1g of Al2O3,We need

. (96÷204)= 0.47g of O2

To produce 95g of Al2O3,We need

. (0.47×95)= 44.706g of O2

.

So, 44.706 grams of oxygen (O2)

are needed to produce 95 grams of

aluminum oxide (Al2O3).