Answer:
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 .
