Calculate the number of milliliters of 0.587 M NaOH required to precipitate all of the Fe3 ions in 197 mL of 0.654 M FeCl3 solution as Fe(OH)3. The equation for the reaction is: FeCl3(aq) 3NaOH(aq) Fe(OH)3(s) 3NaCl(aq)

The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the ions in 197 mL of 0.654 M solution as .

Explanation:

The reaction equation is as follows.

Therefore, moles of are calculated as follows.

Moles = Molarity of   Volume (in L)

= 0.654 M 0.197 L  

= 0.128 mol

Now, according to the given balanced equation 1 mole of reacts with 3 moles of NaOH(aq). Hence, moles of  reacted are calculated as follows.

3 0.128 mol = 0.384 moles of NaOH

As moles of NaOH present are as follows.

Moles of NaOH = Molarity of NaOH Volume (in L)

0.384 mol = 0.587 M 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 ions in 197 mL of 0.654 M solution as .