Question

In an experiment, 3.25 g of Ali are allowed to react with 3.50 g of O2. 4Li + O2 > 2Li2O. Which reactant is the limiting reagent?

Answer 1

Step-by-step explanation:

4Li + O₂ = 2Li₂O.

4 mole 1 mole

3.25 g Li = 3.25 / 7

= .464 moles

3.5 g of O₂ = 3.5 / 32 mole

= .11 mole

.11 mole of O₂ will react with .44 mole of Li as per the equation above

but Li available is .464 which is in excess .

Hence O₂ is in short supply for given mass of Li .

Hence O₂ is the limiting reagent .