Question

Calculate the moles of oxygen that's in excess in the reaction of 1.0.g of litium and 1.5g of oxygen

Answer 1

2Li + O₂ ------> 2LiO

Molar mass of Li= 7.0g/mol

Molar mass of O₂ = 16 x 2 = 32g/mol

Mass of Li= 1.0g

Mass of oxygen= 1.5g

From the balanced equation

2(7.0g) of Li reacts with 32g of oxygen

1.0g of Li will react with 32x1.0/14 = 2.28g of O₂

since only 1.5g of oxygen is available, it is the limiting reagent. Hence, Lithium in this case is the excess reagent.

32g of O₂ reacts with 14g of Li

1.5g of O₂ will react with 14 x 1.5/32

= 0.66g of Li

Mass of Li in excess = 1.0g - 0.66g

= 0.34g

Mole of Li in excess= mass/molar mass

= 0.34/7.0 = 0.5moles