Pb + PbO2 + H2SO4 --> PbSO4 + H2O
Pb(NO3)2 + H2SO4 -> PbSO4 + 2HNO3
mg,pb,au
Should be, assuming ideal reaction....... Pb + H2SO4 --> PbSO4 + H2 Products are lead sulfate and hydrogen gas.
For lead (II) nitrate: H2SO4 + Pb(NO3)2 -----> 2HNO3 + PbSO4 ...................................................(white)
Pb + PbO2 + H2SO4 --> PbSO4 + H2O
Pb(NO3)2 + H2SO4 -> PbSO4 + 2HNO3
The chemical equations are:PbO2 + 2 H2SO4 + Pb(SO4)2 + 2 H2OPbO + H2SO4 = PbSO4 + H2O
the positive plate is Pb the negative is PbO2 and the acid is H2SO4.
mg,pb,au
Reactions of Lead (Pb) with cold, dilute acids are very slow. Some examples are as follows:- 1) Reaction with dil. HCl and gives PbCl2 and H2 gas. Pb + 2HCl ---> PbCl2 + H2 Solid Aqs. Aqs. Gas 2) Reaction with dil. H2SO4 and gives PbSO4 and H2 gas. Pb + H2SO4 ----> PbSO4 + H2 Solid Aqs. Aqs. Gas 3) Reaction with dil. HNO3 and gives Pb(NO3)2 and H2 gas. Pb + 2HNO3 ------> Pb(NO3)2 + H2 Solid Aqs. Aqs. Gas Regards- Ashutosh talktomehiya@gmail.com
Should be, assuming ideal reaction....... Pb + H2SO4 --> PbSO4 + H2 Products are lead sulfate and hydrogen gas.
For lead (II) nitrate: H2SO4 + Pb(NO3)2 -----> 2HNO3 + PbSO4 ...................................................(white)
I believe that there is a typo in the formula, because of the ionization state of the lead. In the reactant PbO2, each oxygen carries a -2 charge, so the Pb must be +4, but in the products, PbSO4 is the problem... The SO4 will carry a -2 charge, which means the lead must now be a +2, If the lead went from +4 to +2, it would have had to get 2 electrons from somewhere.... So, the equation is balanced if it is: PbO2 + 2H2SO4 ---> Pb(SO4)2 + 2H2O or: PbO + H2SO4 ---> PbSO4 + H2O
Let's see. Pb(NO3)2 + H2SO4 --> PbSO4 + 2HNO3 If double displacement, lead sulfate ( a salt ) and nitric acid.
Pb(NO3)2
Assuming the 2 oxidation state of lead. Pb + 2HNO3 --> Pb(NO3)2 + H2