I have an ap chem question?

Question

this was a hw problem i had trouble with

When 200.0 ml of a 0.400 M solution of iron (III) chloride is mixed with 300.0 ml of a 0.700 M solution of sodium hydroxide, 6.43 grams of the precipitate, iron (III) hydroxide, are recovered. (24 points total, 3 points each)

1. Write the balanced net ionic equation for this reaction.

2. What is the limiting reagent?

3. How many grams of precipitate should have been formed theoretically?

4. What was the percent yield for this experiment?

5. What was the molarity of the Fe+3 ion that remained free in the final solution?

6. What was the molarity of the Cl- ion that remained free in the final solution?

7. What was the molarity of the Na+ ion that remained free in the final solution?

8. What was the molarity of the OH- ion that remained free in the final solution?

Answer 1

(1) Fe^+3 + 3OH^-1 --> Fe(OH)3

(2) NaOH is limiting

200.0 mL x 0.400 M = 80.00 mmol Fe^+3

300.0 mL x 0.700 M = 210.0 mmol OH^-1

(3) 7.480 g Fe(OH)3

210 mmol OH^-1 should yield 70 mmol Fe(OH)3

70 mmol x 106.87 g/mol x (1/1000) = 7.480 g

(4) 86%

6.43/7.48 = 0.86

(5) 0.0200 M

80 mmol - 70 mmol = 10 mmol

10 mmol/500 mL = 0.0200 M

(6) 0.480 M

3 x 80 mmol / 500 mL = 0.480 M

(7) 0.420 M

210 mmol/500 mL = 0.420 M

(8) 0.000 M

Assuming pH of 7, [Oh-] = 1x10^-7 M