Calculate the number of milliliters of 0.568 m ba(oh)2 required to precipitate all of the fe3+ ions in 165 ml of 0.663 m fe(no3)3 solution as fe(oh)3. the equation for the reaction is: 2fe(no3)3(aq) + 3ba(oh)2(aq) 2fe(oh)3(s) + 3ba(no3)2(aq)

289 mL Ba(OH)2

Step 1. Write the chemical equation

2Fe(NO3)3 + 3Ba(OH)2 → 2Fe(OH)3 + 3Ba(NO3)2

Step 2. Calculate the moles of Fe(NO3)3

Moles of Fe(NO3)3

= 165 mL Fe(NO3)3 × (0.663 mmol Fe(NO3)3/1 mL Fe(NO3)3) = 109.4 mmol Fe(NO3)3

Step 3. Calculate the moles of Ba(OH)2

Moles of (BaOH)2

= 109.4 mmol Fe(NO3)3× [3 mmol Ba(OH)2]/[2 mmol Fe(NO3)3]

= 164.1 mmol Ba(OH)2

Step 4. Calculate the volume of Ba(OH)2

Volume of Ba(OH)2 = 164.1 mmol Ba(OH)2 ×[ 1 mL Ba(OH)2]/[0.568 mmol Ba(OH)2] = 289 mL Ba(OH)2

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