You wish to prepare 222 grams of 22.1 % FeCl2. You will need grams of iron(II) chloride and mL of water. Assume that the density of water is 1.00 g / ml

I will need 49 grams of iron(II) chloride and 173 mL of water .

Explanation:

222 grams of 22.1 % FeCl₂ will require

222 x 0.221 gram of FeCl₂ and rest will be water .

49 gram of  FeCl₂ and 173 grams of water .

density of water is 1.00 g / mL

volume of water in mL = 173 / 1

= 173 mL

Hence I will need 49 grams of iron(II) chloride and 173 mL of water .

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