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)

Question

Chemistry

Natalee

11 June, 12:19

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)

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Aliana Fernandez · Answer 1

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