Answer: 44.706g Al2O3
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).
