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According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that there is excess FeCl2.

FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)

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The number of moles of Fe(OH)2 that can form is equal to the number of moles of LiOH present in the solution. Since the solution is 0.227 M LiOH, the number of moles of LiOH present is 0.227 moles/L x 175.0 mL = 39.675 moles. Therefore, 39.675 moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution.
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