What is the theoretical yield (in grams) of the precipitate when 22.2 g of iron(

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FeCl2 + 2 NaOH ====> Fe(OH)2 + 2 NaCl Moles = mass/molar mass Moles of NaOH = 10.91/40 = 0.27 mol NaOH Moles of FeCl2 = 22.2/126.75 = 0.18 mol FeCl2 the molar ratio is 1 FeCl2 : 2 NaOH 0.27 mol NaOH/2 = 0.135 mol NaOH 0.18 mol FeCl2/1 = 0.18 mol FeCl2 So NaOH is the limiting reagent 0.27 mol NaOH * (1 mol Fe(OH)2 / 2 mol NaOH)*(89.86 g Fe(OH)2/ 1mol Fe(OH)2) 12.13 g of Fe(OH)2

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