Respuesta :
Answer: The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].
Explanation:
The reaction equation is as follows.
[tex]FeCl_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaCl(aq)[/tex]
Therefore, moles of [tex]Fe(OH)_{3}[/tex] are calculated as follows.
Moles = Molarity of [tex]Fe(OH)_{3}[/tex] [tex]\times[/tex] Volume (in L)
= 0.654 M [tex]\times[/tex] 0.197 L
= 0.128 mol
Now, according to the given balanced equation 1 mole of [tex]FeCl_{3}(aq)[/tex] reacts with 3 moles of NaOH(aq). Hence, moles of [tex]Fe(OH)_{3}[/tex] reacted are calculated as follows.
3 [tex]\times[/tex] 0.128 mol = 0.384 moles of NaOH
As moles of NaOH present are as follows.
Moles of NaOH = Molarity of NaOH [tex]\times[/tex] Volume (in L)
0.384 mol = 0.587 M [tex]\times[/tex] Volume (in L)
Volume (in L) = 0.654 L (1 L = 1000 mL) = 654 mL
Thus, we can conclude that the number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].
