What volume of 0.479 M lithium hyride is required to neutralize 0.627 g of oxalic acid, H2C2O4?

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What volume of 0.479 M lithium hyride is required to neutralize 0.627 g of oxalic acid, H2C2O4?

2007-07-17 04:47:48 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

hydroxide not hyride

2007-07-17 04:49:14 · update #1

1 answers

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

2007-07-17 05:30:04 · answer #1 · answered by ag_iitkgp 7 · 0⤊ 0⤋