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)

0.020 moles of can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of produced:

As the problem says that there are excess of , the limiting reagent is the LiOH.

can be formed

the answer would be a