Respuesta :
Answer:
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 .
