This set of questions covers writing ionic and net ionic equations for precipitation and neutralisation reactions, as well as balancing chemical equations. Below, each part is solved step-by-step.
Question a: Ionic and Net Ionic Equations for Neutralisation
Given equation: 2HCl(aq)+Ba(OH)2(aq)→BaCl2(aq)+2H2O(l)
Ionic Equation
Break all strong electrolytes into ions: 2H+(aq)+2Cl−(aq)+Ba2+(aq)+2OH−(aq)→Ba2+(aq)+2Cl−(aq)+2H2O(l)
Net Ionic Equation
Cancel spectator ions (Ba2+ and Cl−): 2H+(aq)+2OH−(aq)→2H2O(l) Or simplified: H+(aq)+OH−(aq)→H2O(l)
Question b: Ionic and Net Ionic Equations for Precipitation
Given equation: Pb(NO3)2(aq)+(NH4)2SO4(aq)→PbSO4(s)+2NH4NO3(aq)
Ionic Equation
Break all soluble salts into ions: Pb2+(aq)+2NO3−(aq)+2NH4+(aq)+SO42−(aq)→PbSO4(s)+2NH4+(aq)+2NO3−(aq)
Net Ionic Equation
Cancel spectator ions (NH4+ and NO3−): Pb2+(aq)+SO42−(aq)→PbSO4(s)
Question c: Ionic and Net Ionic Equations for Precipitation
Given equation: CuSO4(aq)+2NaOH(aq)→Cu(OH)2(s)+Na2SO4(aq)
Ionic Equation
Break all soluble salts into ions: Cu2+(aq)+SO42−(aq)+2Na+(aq)+2OH−(aq)→Cu(OH)2(s)+2Na+(aq)+SO42−(aq)
Net Ionic Equation
Cancel spectator ions (Na+ and SO42−): Cu2+(aq)+2OH−(aq)→Cu(OH)2(s)
Question 7: Balancing Chemical Equations
i. PCl5(s) + H2O(l) → POCl3(l) + HCl(aq)
Unbalanced: PCl5+H2O→POCl3+HCl
Balance Cl and H: PCl5+H2O→POCl3+2HCl
ii. Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + H2O(l) + NO(g)
Unbalanced: Cu+HNO3→Cu(NO3)2+H2O+NO
Balance: 3Cu+8HNO3→3Cu(NO3)2+4H2O+2NO
iii. (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + H2O(l)
Unbalanced: (NH4)2Cr2O7→Cr2O3+N2+H2O
Balance: (NH4)2Cr2O7→Cr2O3+N2+4H2O
iv. KNO3 + H2CO3 → K2CO3 + HNO3
Unbalanced: KNO3+H2CO3→K2CO3+HNO3
Balance K: 2KNO3+H2CO3→K2CO3+2HNO3
Summary Table:
ReactionIonic EquationNet Ionic Equationa2H++2Cl−+Ba2++2OH−→Ba2++2Cl−+2H2O2H++2OH−→2H2ObPb2++2NO3−+2NH4++SO42−→PbSO4+2NH4++2NO3−Pb2++SO42−→PbSO4cCu2++SO42−+2Na++2OH−→Cu(OH)2+2Na++SO42−Cu2++2OH−→Cu(OH)2Balanced equations for Q7:
- PCl5+H2O→POCl3+2HCl
- 3Cu+8HNO3→3Cu(NO3)2+4H2O+2NO
- (NH4)2Cr2O7→Cr2O3+N2+4H2O
- 2KNO3+H2CO3→K2CO3+2HNO3