[narf66]

Im sppose to balance these equations but they seem darn near impossible;

Pb(CrO4)2+HCL+FeSO4=PbCl2+Cr2(SO4)3+H2O+Fe(SO4)3


HELP please?

[kingchemist]

It is difficult because they involve oxidation and reduction

CrO42- -----> Cr3+
Pb(IV) ------> Pb(II)
Fe(II) -----> Fe(III)

Pb(CrO4)2 ----> Pb2+ + 2Cr3+
Pb(CrO4)2 ----> Pb2+ + 2Cr3+ + 8H2O
Pb(CrO4)2 + 16H+ ----> Pb2+ + 2Cr3+ + 8H2O
Pb(CrO4)2 + 16H+ ----> Pb2+ + 2Cr3+ + 8H2O + 8e- (oxidation)

Fe2+ ----> Fe3+ + e-  (also oxidation)
Something wrong.
Are you sure

You have an error in one formula Fe2(SO4)3 not Fe(SO4)3

However, it is still not possible I think

Could Pb(CrO4)2 be wrong and could it be PbCrO4

Still can't balance it. You can't get enough SO4 ions as you start with 1 x Fe : 1 x SO4 and you end with Fe2(SO4)3 so 1 x Fe : 1.5 SO4 and you also need SO4 to make Cr2(SO4)3

hey

[Hertzinger]

It is difficult because they involve oxidation and reduction

CrO42- -----> Cr3+
Pb(IV) ------> Pb(II)
Fe(II) -----> Fe(III)

Pb(CrO4)2 ----> Pb2+ + 2Cr3+
Pb(CrO4)2 ----> Pb2+ + 2Cr3+ + 8H2O
Pb(CrO4)2 + 16H+ ----> Pb2+ + 2Cr3+ + 8H2O
Pb(CrO4)2 + 16H+ ----> Pb2+ + 2Cr3+ + 8H2O + 8e- (oxidation)

Fe2+ ----> Fe3+ + e-  (also oxidation)
Something wrong.
Are you sure

You have an error in one formula Fe2(SO4)3 not Fe(SO4)3

However, it is still not possible I think

Could Pb(CrO4)2 be wrong and could it be PbCrO4

Still can't balance it. You can't get enough SO4 ions as you start with 1 x Fe : 1 x SO4 and you end with Fe2(SO4)3 so 1 x Fe : 1.5 SO4 and you also need SO4 to make Cr2(SO4)3

He is absolutely right it's redox-radox reaction rates
hey

[valdorod]

Kingchemist is correct, the entered equation cannot be solved nor can it be solved by changing the last term

the most likely equation is probably this one
which means that the first term was entered incorrectly and it is missing a term on the products side.

2 PbCrO4 + 16 HCl + 6 FeSO4 = 2 PbCl2 + Cr2(SO4)3 + 4 FeCl3 + 8 H2O + Fe2(SO4)3


hey