Question

Considered the reaction:

3 NaOH(aq)+ FeCI3(s) →3 NaCI (aq) + Fe(OH)3 (s)

(molar mass: Fe(OH)3=106.84 g/mole; FeCI3=162.19 g/mol

2) what volume of 1.6M NaOH will react with 33.0g of FeCI3

3. What is the concentration of 22.50 mL of NaOH if this was neutralized with 20.50 mL of 0.150 M HCI in the following reaction

Answer 1

33g of Fecl3 in moles=

Moles = mass/rfm

Moles= 33/162= 0.2037

NaOH: FeCl3= 3:1

Moles of NaOH are

(3/4)× 0.2037= 0.1528moles.

Volume that reacted is;

(0.1528/1.6)× 1000= 95.48ml