F.4 Chemistry Acid & Alkalis問題

(a)(i) ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(l)
(a)(ii) ZnO(s) + 2H+ (aq) + 2Cl- (aq) → Zn2+(aq) +2Cl- (aq) + H2O(l)
ZnO(s) + 2H+(aq) → Zn2+(aq) + H2O(l)

(b)(i) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
(b)(ii) CaCO3(s)+2H+(aq) + 2Cl-(aq) → Ca 2+(aq)+ 2Cl-(aq) + CO2(g)+H2O(l)
CaCO3(s)+2H+(aq) → Ca 2+(aq) + CO2(g)+H2O(l)

(c)(i) KHCO3(aq)+HCl(aq) → KCl(aq)+CO2(g)+H2O(l)
(c)(iI) K+HCO3- (aq) + H+(aq) + Cl-(aq) → K+(aq) + Cl-(aq) +CO2(g)+H2O(l)
HCO3 -(aq) + H+(aq) → CO2(g)+H2O(l)

2010-04-08 17:34:09 補充：
When we write an ionic equation, we need to decide which substances exist as mobile ions in the solution. So, we need to know the substances soluble [i.e. (aq)] or not.

2010-04-08 17:38:56 補充：
such as Cu + 2AgNO3 → Cu(NO3)2 + 2Ag, if we know that AgNO3 and Cu(NO3)2 are aq, We can write the equation as Cu(s) + 2Ag+(aq) +2NO3-(aq) → Cu2+ (aq) 2NO3- (aq) + 2Ag(s), then we need to cross out the same sub. in the equation [if aq].

2010-04-08 17:39:53 補充：
finally, we can obtain that the ionic equation: Cu(s) + 2Ag+(aq) → Cu2+ (aq) + 2Ag(s).

For full equation, you should first see what the reaction will occur in the above equation.

Question a(i), neutralization occurs. A salt and water molecules are product.

ZnO (s) + 2HCl (aq) ---> ZnCl2 (aq) + H2O (l)

Question b(i) & c(i), a salt, carbon dioxide gas and water are given as product.

CaCO3 (s) + 2HCl (aq) ---> CaCl2 (aq) + H2O (l) + CO2 (g)

KHCO3 (aq) + HCl (aq) ---> KCl (aq) + H2O (l) + CO2 (g)

Note: Remeber to balance the equation!

___________________________________________________________

It is very easy to write an ionic equation if you know how to write full equation. First of all, you should know which species is REALLY undergoing reaction.

For (a), you should write the ions present in the substance. Since zinc oxide is INSOLUBLE in water, you CANNOT write Zn2+ and O2- in the reaction.

Instead, the reaction should be:

ZnO (s) + H+ (aq) + Cl- (aq) ---> Zn2+ (aq) + Cl- (aq) + H2O (l)

However, it is not yet finished. You have to cancel all species that didn't participate in the whole reaction. (i.e. Cl-)

The answer of a should be:

ZnO (s) + 2H+ (aq) ---> Zn2+ (aq) + H2O (l)


By the same principle, the answer of the rest questions are:

(b) CaCO3 (s) + 2H+ (aq) ---> Ca2+ (aq) + CO2 (g) + H2O (l)

(c) HCO3- (aq) + H+ (aq) ---> H2O (l) + CO2 (g)


Note: Hints are important to you when doing this questions. As the hint tell you that the two substance are insoluble, then you can write the state symbol of them and to determine whether they should be written as ions when writing ionic equation.




Hope my explanations help.

(a)Zinc Oxide+ dilute Hydrochloric Acid
(i)ZnO(s)+2HCl(aq)->ZnCl2(aq)+H2O(l)
(ii)ZnO(s)+2H+(aq)->Zn 2+(aq)+H2O(l)

(b)Calcium Carbonate + dilute hydrochloric Acid
(i)CaCO3(s)+2HCl(aq)->CaCl2(aq)+CO2(g)+H2O(l)
(ii)CaCO3(s)+2H+(aq)->Ca 2+(aq)+CO2(g)+H2O(l)

(c) Potassium Hydrogencarbonate + dilute hydrochloric Acid(
(i)KHCO3(aq)+HCl(aq)->KCl(aq)+CO2(g)+H2O(l)
(ii)HCO3-(aq)->CO2(g)+H2O(l)



2010-04-08 17:10:40 補充：
The hint is useful when we write the state symbol of the chemical.
Moreover, we need to know whether the chemical is soluble or not when writing the ionic equation.

2010-04-08 17:10:48 補充：
It is because when it is soluble, ions are formed. KHCO3(aq) forms K+(aq) and HCO3-(aq)
When it is insoluble, it cannot form ions when being put into water. CaCO3(s) is not written into Ca2+(aq) and CO3 2-(aq) in this case.