You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each of the following reactions: a. 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 .

See explanation

Explanation:

a) the reaction between KI and Pb(NO3)2 occurs as follows;

2KI(aq) + Pb(NO3)2(aq) ---> PbI2(s) + 2KNO3

Hence the yellow precipitate in the reaction is PbI2

b) the reaction between BaCl2 and H2SO4 occurs as follows;

BaCl2(aq) + H2SO4(aq) ---> BaSO4(s) + 2HCl(aq)

Hence the white precipitate in the reaction is BaSO4

c) the reaction between NaOH and FeCl3 is as follows;

3NaOH(aq) + FeCl3(aq) > Fe(OH)3 + 3NaCl(aq)

Hence the brown precipitate is Fe(OH)3

d) the reaction between CuSO4 and NaOH occurs as follows;

CuSO4(aq) + 2NaOH(aq) ---> Cu(OH)2(s) + Na2SO4(aq)

Hence the blue precipitate is Cu(OH)2

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the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.