Answer:
a. Here, the yellow precipitate is lead (II) iodide:
[tex]2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)[/tex]
b. Here the white precipitate is barium sulfate:
[tex]BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)[/tex]
c. Here the brown precipitate is iron (III) hydroxide:
[tex]3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)[/tex]
d. Here, the precipitate is copper (II) hydroxide:
[tex]CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)[/tex]
Explanation:
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In this case, since the precipitation reactions are characterized by the formation of a product which is highly insoluble in water, specially very heavy products having heavy metals at the cations, for each case we proceed as follows:
a. Here, the yellow precipitate is lead (II) iodide:
[tex]2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)[/tex]
b. Here the white precipitate is barium sulfate:
[tex]BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)[/tex]
c. Here the brown precipitate is iron (III) hydroxide:
[tex]3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)[/tex]
d. Here, the precipitate is copper (II) hydroxide:
[tex]CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)[/tex]
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Solid products that are obtained from two aqueous phase reactants is called a precipitate.
A precipitate is formed when a solid product results from the reaction of two aqueous solutions. The precipitate is that solid product which may or may not be colored.
The reaction between KI and Pb(NO3)2 occurs as follows;
2KI(aq) + Pb(NO3)2(aq) --------> PbI2(s) + 2KNO3(aq)
The yellow precipitate is PbI2.
The reaction between BaCl2 and H2SO4 occurs as follows;
BaCl2(aq) + H2SO4(aq) -------> BaSO4(s) + 2HCl(aq)
The white precipitate formed in this reaction BaSO4
The reaction between NaOH and FeCl3 occurs as follows;
3NaOH(aq) + FeCl3(aq) ------> Fe(OH)3(s) + 3NaCl(aq)
The brown precipitate formed in this reaction is Fe(OH)3.
The reaction between CuSO4 and NaOH occurs as follows;
CuSO4(aq) + 2NaOH(aq) ---------> Cu(OH)2(s) + Na2SO4(aq)
The blue precipitate formed in this reaction is Cu(OH)2.
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