Two metals X and Y form the salts XSO₄ and Y₂SO₄ respectively .
The solution of salt XSO₄ is blue in colour and whereas that of Y₂SO₄ is colourless.
When Barium chloride solution is added to XSO₄ solution , then a white precipitate Z is formed along with a salt which turns the solution green.
And when barium chloride solution is added to Y₂SO₄ solution , then the same precipitate Z is formed along with the colourless common salt solution .
a) what could the metals X and Y be ?
b) Write the name and the formula of the salt XSO₄?
c) Write the name and the formula of the salt Y₂SO₄ ?
d) What is the name and formula of precipitate Z ?
e) Write the name and formula of the salt which turns the solution green in the first case
Answers
Answer:
beaker B in which zinc was placed, due to displacement reaction grey deposit forms on the zinc when compound of zinc in solution displaced iron . Whereas, in beaker A in which copper was placed, displacement reaction doesn't occur as copper can't displace iron because it is less reactive than iron.
Detailed Answer [Explanation] :
As per the given information :
Both A and B beakers contain Iron sulphate. \sf { (Fe{SO}_{4}) }(FeSO
4
)
Small piece of copper is placed in the Beaker A.
Small piece of zinc is placed in the Beaker B.
In beaker B , green deposits forms on the zinc but not on the copper.
_______________________________
• In beaker A : Iron Sulphate + Copper
• In beaker B : Iron Sulphate + Zinc
________________________________
Reaction :
Here, displacement reaction will occur in which a more reactive metal displaces a less reactive metal from its compound in solution.
★ \boxed {\sf{ Metal \: A + Salt \: Solution \: of \: B \longrightarrow Salt \: Solution \: of \: A + Metal \: B }}
MetalA+SaltSolutionofB⟶SaltSolutionofA+MetalB
In Beaker A :
\sf { \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Cu}_{Copper} \longrightarrow \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Cu}_{Copper} }
IronSulphate
FeSO
4
+
Copper
Cu
⟶
IronSulphate
FeSO
4
+
Copper
Cu
As copper is less reactive than iron. So, it can't displace iron from its compound in solution. Reactants and product will be same.
In Beaker B:
\sf { \underbrace{FeSO_4}_{Iron \: Sulphate} + \underbrace{Zn}_{Zinc} \longrightarrow \underbrace{ZnSO_4}_{Zinc \: Sulphate} + \underbrace{Fe}_{Iron} }
IronSulphate
FeSO
4
+
Zinc
Zn
⟶
ZincSulphate
ZnSO
4
+
Iron
Fe
Here, zinc is more reactive metal than iron. So, when zinc (Metal A) reacts with Iron sulphate (Salt solution of B) , zinc displaced iron from its compound in solution and forms zinc sulphate (Salt solution of A) and Iron (Metal B).There will be a grey deposit forms on the zinc.
______________________________________
Thus, it can be concluded from these observations that
• Zinc is more reactive than copper, that's why it displaced iron and a grey deposit forms on the zinc or it became zinc sulphate.
• Whereas, in the case of copper , copper couldn't displace iron as it is less reactive than the iron. That's why a grey deposit doesn't form on the copper.u
Answer:
mention pic of question