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


1
Expert's answer
2021-02-09T03:47:34-0500

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS