Question

The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is

Answer 1

Answer:

a. Here, the yellow precipitate is lead (II) iodide:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

b. Here the white precipitate is barium sulfate:

$BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)$

c. Here the brown precipitate is iron (III) hydroxide:

$3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)$

d. Here, the precipitate is copper (II) hydroxide:

$CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)$

Explanation:

Hello.

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:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

b. Here the white precipitate is barium sulfate:

$BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)$

c. Here the brown precipitate is iron (III) hydroxide:

$3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)$

d. Here, the precipitate is copper (II) hydroxide:

$CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)$

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