Write a balanced chemical equation for the reaction of aqueous potassium hydroxide with aqueous iron(III) chloride to form solid iron(III) hydroxide and aqueous potassium chloride. Express your answer as a chemical equation including phases.

The answer to your question is

             3KOH (aq)  +  FeCl₃(aq)   ⇒    Fe(OH)₃ (s)  +   3KCl (aq)

Explanation:

Data

Potassium hydroxide = KOH

Iron (III) chloride = FeCl₃

Iron (III) hydroxide = Fe(OH)₃

Potassium chloride = KCl

- Write the chemical equation

               KOH (aq)  +  FeCl₃(aq)   ⇒    Fe(OH)₃ (s)  +   KCl (aq)

                  Reactants              Elements        Products

                        1                             K                       1

                        1                             Fe                     1

                        3                            Cl                      1

                        1                              H                      3

                        1                              O                      3

This reaction is unbalanced

- Balancing the reaction

              3KOH (aq)  +  FeCl₃(aq)   ⇒    Fe(OH)₃ (s)  +   3KCl (aq)

                  Reactants              Elements        Products

                        3                             K                      3

                        1                             Fe                     1

                        3                            Cl                      3

                        3                              H                      3

                        3                              O                      3

This reaction is now balanced

a plasma is a warm ionized gasoline consisting of approximately equal numbers of definitely charged ions and negatively charged electrons. the characteristics of plasmas are notably unique from the ones of ordinary neutral gases in order that plasmas are considered a awesome "fourth kingdom of remember."

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol