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

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