How many grams of solid NaCN have to be added to 1.4 L of water to dissolve 0.11 mol of Fe(OH)3in the form of Fe(CN)63−? (For simplicity, ignore the reaction of CN−ion with water.) Solution 3 moles of NaCN dissolve 1 mole of Fe(OH)3 3*.11 = .33 moles NaCN will dissolve .11 mole of Fe(OH)3 wt reqd. = 0.33*(23+12+14) w = 16.17 g.