Q)A certain quantity of electricity liberates 9g of aluminiun. The volume of oxygen liberated at r.t.p from dilute sulphuric acid by the same quantity of electricity is:

A)6 dm^3
B)8 dm^3
C)12 dm^3
D)24 dm^3

Question

Q)A certain quantity of electricity liberates 9g of aluminiun. The volume of oxygen liberated at r.t.p from dilute sulphuric acid by the same quantity of electricity is:

A)6 dm^3
B)8 dm^3
C)12 dm^3
D)24 dm^3

anonymous · Answer

electrolysis = decomposition

$$2Al_2O_3\rightarrow4Al+3O_2$$
find the no of moles of AL= 1/3 moles

4AL : 3O2
1/3 : x
x =  1/4 which is the number of moles of O2

then multiply with 24dm^3 =  6 dm^3

anonymous · Answer

Where does dilue sulphuric acid come in this? >_<

anonymous · Answer

i think its 6 whats the answer?

anonymous · Answer

6 dm^3 is the answer. But why work it out by the decompoition of Al2O3 and not by H2SO4?

anonymous · Answer

coz even in the decomposition of H2SO4


ionspresent in it = H+/SO4-
                           H+/OH-
O2 is dissipated in the same amount, coz the same amount current  is used...and u can see the conc of O2 is simmilar to that of aluminum oxide...but if u take it as hydrogen then it would double...

anonymous · Answer

Okay, thanks!