I would say that you could classify it as a precipitation reaction due to the fact that lead(II) sulfate is formed which is an insoluble compound becoming the precipitate during the reaction.
I would say that you could classify it as a precipitation reaction due to the fact that lead(II) sulfate is formed which is an insoluble compound becoming the precipitate during the reaction.
In aqeous solution lead nitrate is dissociated into Pb2+ and NO3-, which helps in consequent formation of lead sulphate, also lead sulphate is dissolved in concentrated sulfuric acid
In aqeous solution lead nitrate is dissociated into Pb2+ and NO3-, which helps in consequent formation of lead sulphate, also lead sulphate is dissolved in concentrated sulfuric acid
Mr. Melaika is correct. You need a source of sulfate ions, like sulfuric acid, so …
PbCO3 + H2SO4 = PbSO4 + CO2(g) + H2O(l)
Change in Free Energy: ΔG(20C) = -158.6kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -99.6kJ (negative, so the reaction is exothermic)
Mr. Melaika is correct. You need a source of sulfate ions, like sulfuric acid, so …
PbCO3 + H2SO4 = PbSO4 + CO2(g) + H2O(l)
Change in Free Energy: ΔG(20C) = -158.6kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -99.6kJ (negative, so the reaction is exothermic)
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I would say that you could classify it as a precipitation reaction due to the fact that lead(II) sulfate is formed which is an insoluble compound becoming the precipitate during the reaction.
I would say that you could classify it as a precipitation reaction due to the fact that lead(II) sulfate is formed which is an insoluble compound becoming the precipitate during the reaction.
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This is an exchange reaction followed by a precipitate forming:
Pb(NO3)2 + H2SO4 = PbSO4↓ + 2HNO3. Lead sulfate is a white precipitate
This is an exchange reaction followed by a precipitate forming:
Pb(NO3)2 + H2SO4 = PbSO4↓ + 2HNO3. Lead sulfate is a white precipitate
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In aqeous solution lead nitrate is dissociated into Pb2+ and NO3-, which helps in consequent formation of lead sulphate, also lead sulphate is dissolved in concentrated sulfuric acid
PbSO4(S)+ H2SO4(conc)= Pb(HSO4)2(aq)
In aqeous solution lead nitrate is dissociated into Pb2+ and NO3-, which helps in consequent formation of lead sulphate, also lead sulphate is dissolved in concentrated sulfuric acid
PbSO4(S)+ H2SO4(conc)= Pb(HSO4)2(aq)
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See there can be 2 products
1: If Temperature is less than 200 ° C
NaNO3 + H2SO4 →NaHSO4 + HNO3
( Process of nitric acid preparation)
2: If Temperature is more than 200°C
NaNO3 + H2SO4 → Na2SO4 + HNO3
Na2SO4 sticks to glass vessel , difficult to remove as it sticks to glass , it may cause the glass to crack , and there unnecessary wastage of fuel …
:) Hope you like my answer….
See there can be 2 products
1: If Temperature is less than 200 ° C
NaNO3 + H2SO4 →NaHSO4 + HNO3
( Process of nitric acid preparation)
2: If Temperature is more than 200°C
NaNO3 + H2SO4 → Na2SO4 + HNO3
Na2SO4 sticks to glass vessel , difficult to remove as it sticks to glass , it may cause the glass to crack , and there unnecessary wastage of fuel …
:) Hope you like my answer….
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(1) Pb(NO3)2 + H2SO4 = PbSO4 + 2HNO3
ΔG(20C) = -20.3kJ (negative, so the reaction runs)
ΔH(20C) = +2.4kJ (positive, so the reaction is endothermic – needs heat applied)
(2) PbSO4 + 2HNO3 = Pb(NO3)2 + H2SO4
ΔG(20C) = +20.3kJ (positive, so the reaction DOES NOT RUN run)
(1) Pb(NO3)2 + H2SO4 = PbSO4 + 2HNO3
ΔG(20C) = -20.3kJ (negative, so the reaction runs)
ΔH(20C) = +2.4kJ (positive, so the reaction is endothermic – needs heat applied)
(2) PbSO4 + 2HNO3 = Pb(NO3)2 + H2SO4
ΔG(20C) = +20.3kJ (positive, so the reaction DOES NOT RUN run)
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There won’t be any meaningful reaction taking place between the two if the acid is dilute, even at high temperatures.
But with the hot and concentrated acid, the products will be copper sulphate and nitric acid.
Cu(NO3)2 + H2SO4 (hot,conc.) = CuSO4 + 2HNO3
Sulphuric acid displces the weaker and more volatile nitric acid from the latter’s salt in this reaction.
There won’t be any meaningful reaction taking place between the two if the acid is dilute, even at high temperatures.
But with the hot and concentrated acid, the products will be copper sulphate and nitric acid.
Cu(NO3)2 + H2SO4 (hot,conc.) = CuSO4 + 2HNO3
Sulphuric acid displces the weaker and more volatile nitric acid from the latter’s salt in this reaction.
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Start with the “molecular” equation with states.
H2SO4(aq_ + Pb(NO3)2 (aq) → PbSO4(s) + 2HNO3(aq)
To write the ionic equation , separate all compounds that are (aq) into ions
2H+(aq) + SO4 2-(aq) + Pb2+(aq) + 2NO3-(aq) → PbSO4(s) + 2H+(aq) + 2NO3-(aq)
And if you want the net ionic equation:
Examine the above ionic equation and delete anything that is identical on both sides of →. In this case: H+ , NO3-
Net ionic equation
SO4 2-(aq) + Pb 2+(aq) → PbSO4(s)
Start with the “molecular” equation with states.
H2SO4(aq_ + Pb(NO3)2 (aq) → PbSO4(s) + 2HNO3(aq)
To write the ionic equation , separate all compounds that are (aq) into ions
2H+(aq) + SO4 2-(aq) + Pb2+(aq) + 2NO3-(aq) → PbSO4(s) + 2H+(aq) + 2NO3-(aq)
And if you want the net ionic equation:
Examine the above ionic equation and delete anything that is identical on both sides of →. In this case: H+ , NO3-
Net ionic equation
SO4 2-(aq) + Pb 2+(aq) → PbSO4(s)
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