According to the reaction below, how many moles of Fe(OH)2(s) can be produced from 150. mL of 0.058 M LiOH(aq)? Assume that there is excess FeCl2(aq).


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

A)0.40 moles Fe(OH)2

B)0.0087 moles Fe(OH)2

C)0.0044 moles Fe(OH)2

D)0.0174 moles Fe(OH)2

E)0.78 moles Fe(OH)2