Question

Nandita mixed 2 solutions X and y. she recorded that following observations and conclusion in her notebook.
A- yellow precipitate is formed
B- it's a double displacement reaction
the solution x and y are
(A) pb(no3)2,kI
(B) AgNO3,NaCL
(C)Na2SO4,BaCL2
(D)FeCL3,NH4OH

Answer 1

Answer:

option:c is the answer

Step-by-step explanation:

BaCl2 when combined with Na2SO4 yellow ppt is observed

AgNO3 when reacts with NaCl gives white precipitation

Answer 2

The solution $x\ and\ y$ are $Pb(NO_{3} )_{2}\ and\ KI$ respectively and option (A) is correct.

Step-by-step explanation:

Given:

Nandita mixed 2 solutions  $x\ and\ y$ .

A- yellow precipitate is formed.

B- it's a double displacement reaction.

To Find:

The solution $x\ and\ y$.

Concept Used:

Double Displacement Reaction

Double displacement reactions occur when a part of two ionic compounds is exchanged and makes two new components.

Double displacement reactions take place generally in aqueous solutions where the ions precipitate and exchange of ions takes place.

Solution:

Solutions of$Pb(NO_{3} )_{2}$ (lead nitrate) ($x$)and $KI$( potassium iodide) ($y$) are mixed.  $PbI_{2}$ (lead iodide) and $KNO_{3}$ (Potassium nitrate) are formed.

The yellow precipitate is of $PbI_{2}$ lead iodide.

$Pb(NO_{3} )_{2} _{(aq.)} +KI _{(aq.)} \rightarrow PbI_{2} _{(solid)} + 2KNO_{3}_{(aq.)}$

Thus, the solution $x\ and\ y$ are $Pb(NO_{3} )_{2}\ and\ KI$ respectively and option (A) is correct.

3SPJ3