What volume of 0.479 M lithium hydroxide is required to neutralize 0.627g of oxalic acid, H2C2O4?
Answer:
The balanced equation gives you the answer...all you need to do is calculate the MW of oxalic acid.
MW=90.0 so you have 0.627/90.0 moles of oxalic acid
0.627/90.0=0.006964 moles
You need 2 moles of LiOH to neutralize 1 mole of oxalic acid.
2x0.006964=0.01393 moles of LiOH (13.93 millimoles)
mL x M = millimoles
mL x 0.479 = 13.93
mL = 13.93/0.479 = 29.08 mL of LiOH
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