Question

27g of Al willl react completely with how much mass of Oz to produce AL2O3

Answer 1

27 g of Al= 1 mole of Al
Balance this equation
4Al + 3O2 → 2Al2O3
Mole ratio
Al : O2
4 : 3
1 : 3/4
Mass of O2 = 3/4 × 32 = 24g

Answer 2

4Al + 3O₂ ---> 2Al₂O₃

4mol + 3 mol ---> 2mol

∴ 4 moles of Al reacts with 3 moles of Oxygen to form 2 moles of Al₂O₃

Now, Moles = weight / Molecular mass
Weight = Moles * Molecular mass

For Al,

Weight = 4 * 27 = 108 grams

For O₂,
Weight = 3 * 32 = 96 grams

For Al₂O₃,
Weight = 2 * 102 = 204 grams


4Al + 3O₂ ---> 2Al₂O₃

108 g + 96 g ---> 204 g

Now,

108 g Al ---> 96g O₂
27 g Al ---> x g O₂

x * 108 = 96 * 27
x = 2592 / 108
x = 24 g O₂

Now, Same way

108 g Al ---> 204 g Al₂O₃
27 g Al ---> y g Al₂O₃

∴ y * 108 = 204 *27
∴ y = 5508/108
∴ y = 51 g Al₂O₃

Answer:- 27 g Al reacts completly with 24 g of O₂ gives 51 g Al₂O₃