chemistry stoichiometry

Question

Approximately (3.230x10^0) g of lead(II) acetate, Pb(CH₃COO)_2, is weighed out and placed in a (2.85x10^2) mL solution of 0.150-mol/L sodium hydroxide. Calculate the mass of precipitate produced in this reaction. 

Note: Your answer is assumed to be reduced to the highest power possible.

Answer 1

First and foremost is this the accurate weight of Lead acetate weighted out? Because 3.230x10^0 is really just 3.230g. Part of the power rules and functions is that any number raised to the 0 power is 1. Then that number 1 multiplied by another number is just the other number. So if this is just another way of saying 3.230g, we can proceed with the reaction understanding that it is 3.230g of Lead acetate or Pb(CH3COO)2 given in the problem.

So you would need to write a reaction of how this proceeds and use the solubility rules to figure out the precipitate.

lead II acetate is reacting with sodium hydroxide:

According to the solubility rules acetates are soluble in aqueous solutions and salts containing Lead are insoluble. Hydrxides are insoluble. But you must apply the solubiity rules from top down, meaning the rules that are up at the top have precedence over the rules that generate thereafter, further down on the list. So the acetate rule is most paramount and top priority it says acetates are soluble. So the precipitate formed would involve Pb moving over with something else.

Then next down on the list of the solubility rules state that salts containing Lead are insoluble. So Lead will break off from acetate and cling to hydroxide.

The hydroxide rule is even further down on the list so it does not take as much precedence as the two previous rules. The hydroxide rule says hydroxides OH- are insoluble in aqueous solutions.

So your reaction will most likely look like this:

Pb(CH3COO)2(aq) + NaOH(aq) ===> Pb(OH)2(s) + NaCH3COO(aq)

The precipitate is Lead Hydroxide, Pb(OH)2

Then make sure the equation is balanced

Pb(CH3COO)2(aq) + 2NaOH(aq) ===> Pb(OH)2(s) + 2NaCH3COO(aq)

with the new coeffecients in front of each species in the reaction you now have a balanced equation.

Now take the grams given for Pb(CH3COO)2 and express this as molarity so you can see now many moles are reflected for 1 L.

Pb 207.2 +

C 12.01 x 2= 24.02

H 1.01 x 3 = 3.03

O 16.00 x2= 32

multiply the organic aspect of Pb(CH3OO)2 by 2 because you have 2 acetate anions

atomic weight of Pb(CH3COO)2 = 325.3 g/mol

You now need mol/L

3.230 g Pb(CH3COO)2 x1mol/325.3g grams will cancel grams top to bottom and you will have moles.

0.009929296 mol of Pb(CH3COO)2

now take the solution given at 2.85x10^2 mL and express this as Liters

2.85x10^2 mL x 1L/1000mL = 0.285 L

make your Molarity for Pb(CH3COO)2

0.009929296 mol/0.285L = 0.034839635 mol/L

Now use the balanced equation and figure out how much of the precipitate will be yieled.

Because you have moles for each of the reactants.

In the solution you have:

0.034839635 mol of Pb(CH3COO)2 and 0.150 mol for 1L of the solution NaOH, how much of the Pb(OH)2(s) will be yielded?

According to the equation of the reaction

Pb(CH3COO)2(aq) + 2NaOH(aq) ===> Pb(OH)2(s) + 2NaCH3COO(aq)

check for limiting reactant again by doing the mol or gram to coeffecient ratio (as long as both reactants are in the same units).

Which ever number is smaller is the limiting reactant

Pb(CH3COO)2 0.034839635 mol/ 1 = 0.034839635

NaOH 0.150/2 = 0.075

So Pb(CH3COO)2 is the limiting reactant it is smaller at 0.034839635 compared to 0.075 for NaOH

Now we will proceed the reaction with Pb(CH3COO)2

According to the equation 1 mol of Pb(CH3COO)2 goes to 1 mol of Pb(OH)2

So if 0.034839635 mol of Pb(CH3COO)2 are in solution, this would result in 0.034839635 mol of Pb(OH)2(s) precipitate according to the equation because it is a 1:1 ratio

So you would have yielded 0.034839635 mol of Pb(OH)2(s), written in scientific notation and rounded to the least significant figures 3.48x10^-2 mol because it is a small number it is raised to the -2.

We have expressed it this way as moles, we can express it as grams as well. All you have to do is use the atomic weight of Pb(OH)2 and translate the moles to grams. You would have:

Pb 207.2

O 16.00 x 2 = 32

H 1.01 x2 =2.02

241.22g/mol x 0.034839635 mol moles will cancel moles top to bottom and you will have grams. 8.40 g or if expressed in scientific notation and proper sig figs, 8.40x10^0 g which is just the same as 8.4g of the precipitate because any number raised to 0 is 1 then multiplied by that number is just that number.