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.

it is the most commonly accepted one

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol