Question

# When you mix 30.0mL of 0.246M FeCl2 with 50.0mL of 0.145M NaOH, a precipitate will form...

When you mix 30.0mL of 0.246M FeCl2 with 50.0mL of 0.145M NaOH, a precipitate will form FeCl2(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!!

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

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