Question

When you mix 30.0mL of 0.246M FeCl₂ with 50.0mL of 0.145M NaOH, a precipitate will form FeCl_2(aq) + 2NaOH_(aq) ---> 2NaCl_(aq) + Fe(OH)_2(s) A. What is the maximum mass of precipitate formed?

B. How many grams of precipitate can be obtaines if percent yeild is 85.0%

C. What is the molarity of Cl^- after the reaction Please explain in steps!! Thank you!!

Answer 1

FeCl2_(aq) + 2NaOH_(aq) --------> 2 NaCl_(aq)   + Fe(OH)2_(s)   

A) moles of FeCl2 = 0.246 x 30/1000 = 0.00738

moles of NaOH = 0.145 x 50 /1000 = 0.00725

NaOH is limiting reagent

0.00725 moles NaOH forms 0.00725 moles Fe(OH)2

convert moles to mass

mass = 0.00725 x 89.86 = 0.65 g

B) % yield = (actual mass / therotical mass) x 100

85 = (actual mass / 0.65) x100

actual mass / 0.65 = 0.85

actual mass = 0.5525 g

C) total volume = 30 + 50 = 80 mL

moles of Cl^- = 0.00738 x 2 = 0.01476

[Cl^-] = 0.01476 / 0.08 = 0.1845 M