Question

Add the physical state for each substance in the following aqueous reactions. Then write and balance a net ionic equation for each reaction. (a) Cr2(SO4)3 + KOH → Cr(OH)3 + K2SO4 (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) AP 17 (b) K2CrO4 + PbCl2 → KCl + PbCrO4 (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) AP 34 (c) Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2 (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) (aq) (s) (g) (l) AP 55

Answer 1

Cr2(SO4)3(aq) +6 KOH(aq) → 2Cr(OH)3(s) +3 K2SO4(aq) is the balanced molecular equation
let us split this in to ionic form to get
complete ionic equation:
2Cr2+(aq) + 3SO42-(aq) + 6K+(aq) + 6OH-(aq) ---> 2Cr(OH)3(s) + 6K+(aq) + 3SO42-(aq)

Now cancel the spectator ions[K+, SO42-,] on both side of the equation we get
Net ionic equation:

2Cr2+(aq) + 6OH-(aq) --->2Cr(OH)3(s)
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K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s). is the molecular equation

complete ionic equation:
2K+(aq) +CrO42-(aq) + Pb2+(aq) + 2Cl-(aq) ---> 2K+(aq) +2Cl-(aq) + PbCrO4(s).
After cancelling spectator ions[K=, Cl-] we get

Net ionic equation:

CrO42-(aq) + Pb2+(aq) -->PbCrO4(s).
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molecular equation is given below
Na2SO3(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + SO2(g)
complete ionic equation:
2Na+(aq) + So32-(aq) + 2H+(aq) + SO42-(aq) ---> 2Na+(aq) +SO42-(aq) +H2O(l) + SO2(g)

Net ionic equation:

SO32-(aq) + 2H+(aq) ----> H2O(l) + SO2(g)
**********************************************
Thank you