Question

how many moles of al2o3 is formed when 5.4g of al and 3.2g of o2

Answer 1

54 g of al requires 48g of o

48g of o requires 54 g of all

3.2g requires ?

(54×3.2)/48

3.6 g of al is required.

3.6+4.8=8.4 g of al203 is formed.