question on balancing equations... (1 Viewer)

[martin310015]

im having problem of balancing this...
1.excess dilute hydrochloric acid added potassium carbonate?
thanx

 

[CM_Tutor]

2HCl_(aq) + K₂CO_{3 (s or aq)} ---> 2KCl_(aq) + CO_{2 (g)} + H₂O_(l)

 

[martin310015]

thanx cm tutor i got a few more question.
1. write equations for the reaction taking place when carbon dioxide is dissolved in aqueous ammonia solution..
2.y can ammonia dissovle more carbon dioxide then pure water

 

[CM_Tutor]

Start with the usual equations for dissolution of carbon dioxide in water:

CO_{2 (g)} <---> CO_{2 (aq)}

CO_{2 (aq)} + H₂O_(l) <---> H₂CO_{3 (aq)}

H₂CO_{3 (aq)} + H₂O_(l) <---> H₃O⁺_(aq) + HCO₃^-_(aq)

In addition, the carbonic acid can react directly with the ammonia present

NH_{3 (aq)} + H₂CO_{3 (aq)} <---> NH₄⁺_(aq) + HCO₃^-_(aq)

And also, the ammonia will react with the hydronium produced by the carbonic acid:

NH_{3 (aq)} + H₃O⁺_(aq) <---> NH₄⁺_(aq) + H₂O_(l)

More CO₂ will dissolve in ammonia because these last two reactions cause the second and third reactions to move right (by applying Le Chatelier's Principle), and thus the first reaction also moves right (Le Chatelier's Principle), thereby increasing the solubility of carbon dioxide.

 

[xiao1985]

sigh~~~ ops... iguess i was a step late... =(

CO₂ (g) <--> CO ₂ (l)
CO₂ (l) + H₂O (l)<--> H<SUB>2</SUB>CO<SUB>3</SUB> (aq)

H<SUB>2</SUB>CO<SUB>3</SUB> (aq) <--> H ⁺(aq) + HCO₃^- (aq)

H⁺ (aq) + NH₃ (aq) <--> NH₄⁺ (aq)

basically, in ammonia you have the last equilibrium in place as well... the last equilibrium will cause some H⁺ ions be used up... therefore it drives the equilibrium in eqn 3 to further right to produce more H⁺ to compensate the change... as a result, more H₂CO₃ will be used, hence more CO₂ are produced to counteract the change...