Question #e4203

1 Answer
Apr 9, 2015

I suppose that there are many combinations in respect to what solutions A, B, C, D, E are. (Please correct me if I am incorrect).

#A= KOH#
#B= NaNO_3#
#C= Zn(NO_3)_2#
#D= HCl#
#E= Na_2CO_3#

A + B: #KOH (aq)#+ #NaNO_3##(aq)# --- > #KNO_3 (aq)# + #NaOH (aq)# (No precipitate reaction)

A+C: #2KOH (aq)# + #Zn(NO_3)_2 (aq)# --> #2KNO_3 (aq) + Zn(OH)_2 (s)#

A+D: #KOH(aq)# + #HCl (aq)# ---> #KCl (aq) + H_2O (l)# (No precipitate reaction)

A+E: #2KOH (aq) + Na_2CO_3 (aq)# ---> #K_2CO_3 (aq) + 2NaOH (aq)#
(No precipitate reaction)

B+C: #NaNO_3 (aq) + Zn(NO_3)_2 (aq) #---> No displacement reaction and no precipitate reaction

B+D: #NaNO_3 (aq) + HCl (aq) # ---> #HNO_3 (aq) + NaCl (aq)#
(No precipitate reaction)

B+E: #NaNO_3 (aq) + Na_2CO_3 (aq)# ---> No displacement reaction and no precipitate reaction

C+D: #Zn(NO_3)_2 (aq) + 2HCl (aq)# ---> #ZnCl_2 (aq) + 2HNO_3 (aq)# (No precipitate reaction)

C+E: #Zn(NO_3)_2 (aq) + Na_2CO_3 (aq)# ---> #2NaNO_3 (aq) + ZnCO_3 (s)#

D+E: #2HCl (aq) + Na_2CO_3 (aq)# ---> #2NaCl (aq) + H_2O (l) + CO_2 (g)# (gas produced)