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.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

m=9

step-by-step explanation: