Answer to Question #161806 in Inorganic Chemistry for PaulB

Question #161806

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

Expert's answer

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

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