What volume of 0.479 M lithium hyride is required to neutralize 0.627 g of oxalic acid, H2C2O4?
Answer:
H2C2O4 + 2 LiH -> Li2C2O4 + 2H2
Molar mass of H2C2O4 = 90 g
Molar mass of LiH = 8 g
0.627 g will take up 0.627*16/90 g of LiH
That is 0.627/45 moles of LiH
or 0.627/(0.479*45) l of the soln. of LiH
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: