Answer to Question #347612 in General Chemistry for Jake

Question

1. If 1.3 L of 3.0M potassium hydroxide completely neutralizes 0.7 L of carbonic acid

(H2CO3), what is the concentration of the carbonic acid? 2KOH+H,CO, K,CO,+2H₂O

2. How many moles of HCl are present in 0.07 L of 6 M HCI? 3. I add 7 mL of 2 M NaOH to 50 mL of water. What is the new concentration?

Rates of Reactions-Le Chatlier's Principle

2KOH+H₂CO,→K.CO,+2H2O overall energy of reaction (AH)=-683.84 kJ/mol 1. Describe 2 ways you could increase the amount of products of this reaction at

equilibrium. 2. Describe 2 ways you could increase the amount of reactants of the reaction at

equilibrium.

Answer 1

1.

n (KOH)= C_M * V = 1,3 * 3 = 3,9 mol

2KOH + H₂CO₃ = K₂CO₃ + 2H₂O

2mol KOH — 1mol H₂CO₃

3,9mol KOH — X

X = 3,9 * 1 / 2 = 1,95 mol H₂CO₃

C_M(H₂CO₃) = n / V = 1,95 / 0,7 = 2,786 mol/l

Answer: 2,786 mol/l H₂CO₃

2.

n(HCl)= C_M * V = 6 * 0,07 = 0,42 mol

Answer: 0,42 mol HCl

3.

C_M (old) = 2 mol/l

V (old) = 7 ml = 0,007 l

V (new) = 57 ml = 0,057 l

n(NaOH)= C_M * V = 2 * 0,007 = 0,014 mol

C_M(new) = n / V(new) = 0,014 / 0,057 = =0,2456 mol/l

Answer: 0,2456 M NaOH

2KOH + H₂CO₃ = K₂CO₃ + 2H₂O + 683,84 kJ

1.

Increasing the reaction products is to shift the equilibrium of the chemical reaction to the right. For this:

a) Since the reaction is exothermic, it must be cooled. 

b) water or K2CO3 should be removed from the reaction. 

c) KOH or H2CO3 should be added.

2.

increasing the amount of reactants shifts the equilibrium of this chemical reaction to the left. For this:

a) Since the reaction is exothermic, it must be heating. 

b) K2CO3 or H2O should be added.