Predict the products of each reaction and write balanced complete ionic and net ionic equations for each. If no reaction occurs, write NO REACTION.
(a) \(\mathrm{BaS}(a q)+\mathrm{NH}_{4} \mathrm{Cl}(a q) \rightarrow\)
(b) \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KCl}(a q) \rightarrow\)
(c) \(\mathrm{KHSO}_{3}(a q)+\mathrm{HNO}_{3}(a q) \rightarrow\)
(d) \(\mathrm{MnCl}_{3}(a q)+\mathrm{K}_{3} \mathrm{PO}_{4}(a q) \rightarrow\)
Equation Transcription:
Text Transcription:
BaS(aq) + NH_4 Cl(aq)
NaC_2 H_3 O_2 (aq) + KCl(aq)
KHSO_3 (aq) + HNO_3 (aq)
MnCl_3 (aq) + K_3 PO_4 (aq)
Solution 97P
(a) BaS(aq) + 2NH4Cl(aq) →BaCl2(aq)+(NH4)2S(aq)
The ions remain in aqueous form.So we cannot have a net ionic equation in this case
(b) Na(CH3COO)2(aq) + KCl(aq) →K(CH3COO)2(aq) +NaCl(aq)
The ions remain in aqueos form.So we cannot have a net ionic equation in this case
(c) KHSO3(aq) + HNO3(aq) →KNO3(aq)+SO2(g)+H2O(l)
H+ (aq) + NO3 -(aq) + K+ (aq) +HSO3 -(aq) ---> SO2 (g) + H2O(l) + NO3 -(aq) +K+(aq)
cancelling the common spectator ions on both sides we get ,
H+(aq) + HSO3-(aq) ---> SO2(g) + H2O(l)
(d) MnCl3(aq) + K3PO4(aq) →MnPO4 (s)+3KCl(aq)
Mn3+ (aq)+ 3Cl- (aq)+ 3K+(aq) + PO4 (aq)3- ---> MnPO4 (s) + 3Cl-(aq) + 3K+(aq)
Cancelling the common spectator ions on both sides we get
Mn3+ (aq) + PO4 3-(aq) ---> MnPO4(s)
