Question

I.

                                   3NaOH(aq) + 3FeCl₃                          NaCl (aq) + 3Fe(OH)₃ (s)

1. What mass of 3Fe(OH)₃ would be produced if 250 mL of 0.250M NaOH were added to a solution containing excess FeCl₃?

2. What volume of 2M NaOH will react with 4.0 g of FeCl₃?

3. What is the concentration of 20.50 mL of NaOH if this was neutralized with 20.0 mL of 0.150 M HCl in the reaction:

NaOH + HCl                       NaCl + HOH

Answer 1

1)

The equation is incorrect. The correct one is:

3NaOH(aq) + FeCl₃(aq) --> 3NaCl(aq) + Fe(OH)₃(s)

$m\ Fe(OH)_3=0.250\ L\ NaOH\times\frac{0.250\ mol\ NaOH}{1\ L\ NaOH}\times\frac{1\ mol\ Fe(OH)_3}{3\ mol\ NaOH}\times\frac{106.87\ g\ Fe(OH)_3}{1\ mol\ Fe(OH)_3}=2.23\ g$

Answer: 2.23 g

2)

$V\ NaOH=4.0\ g\ FeCl_3\times\frac{1\ mol\ FeCl_3}{162.21\ g\ FeCl_3}\times\frac{3\ mol\ NaOH}{1\ mol\ FeCl_3}\times\frac{1\ L\ NaOH}{2\ mol\ NaOH}=0.037\ L=37\ mL$

Answer: 37 mL

3)

$C_{NaOH}V_{NaOH}=C_{HCl}V_{HCl}$

$C_{NaOH}=\frac{C_{HCl}V_{HCl}}{V_{NaOH}}=\frac{0.150\ M\times20.0\ mL}{20.50\ mL}=0.146\ M$

Answer: 0.146 M