Question

How many atoms of aluminum with 48g of oxygen to form A1203

Answer 1

Answer:

Explanation:

+

3moles

3O

2

​

​

⟶

2moles

2Al

2

​

O

3

​

​

3 moles of O

2

​

combine with 4 moles of Al

∴1.5 moles of O

2

​

will combine with x mol of Al

x=

3

4

​

×1.5=2 mol

Thus mass of Al used in the reaction=2×27 g=54 g