If the initial concentrations are {Pb(CH3CO2)2} = 0.165 M and {NH4I} = 0.466 M, how many grams of PbI2 will precipitate out of solution?

76.065 g of PbI2.

Explanation:

Equation of the reaction:

Pb(CH3CO2)2(aq) + 2NH4I --> PbI2(s) + 2NH4(CH3CO2)(aq)

Given:

Pb(CH3CO2)2 = 0.165 M

NH4I = 0.466 M

Since molar concentration = number of moles/volume

Calculating the number of moles of the limiting reagent,

0.165 moles of Pb(CH3CO2)2 in 0.466 moles of NH4I all in the same volume of solution and by stoichiometry, since 1 mole of Pb(CH3CO2)2 reacts with 2 moles of NH4I.

Therefore, Pb(CH3CO2)2 is the limiting reagent.

Using stoichiometry, 1 mole of Pb(CH3CO2)2 reacted to form 1 mole of PbI2 precipitate.

Number of moles of PbI2 = 0.165 moles

Molar mass of PbI2 = 207 + (127 * 2)

= 461 g/mol

Mass = molar mass * number of moles

= 461 * 0.165

= 76.065 g of PbI2.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

qualitative observation deals with data that can be observed with our senses: sight, smell, touch, taste, and hearing. they do not involve measurements or numbers. for instance, colors, shapes, and textures of objects are all qualitative observations.