a large 1.5 x 10^6liter tank contains water polluted with lead (II) nitrate at a molarity concentraion of

2.5 x 10^-6 M

to the tank, you add 59.7 liters of 0.075 M potasium chloride solution.

did you add enough potassium chloride to remove all the lead(II) ions out of the solution??

I didn't work the problem but the answer is no. PbCl2 is relatively soluble and a saturated soln of PbCl2 still contains fairly large concns of Pb.

I think, however, that you are supposed to write and balance the equation, then determine mols Pb in the tank. That will be M x L = ?
Then determine mols KCl you added, go through the stoichiometry and see if there is any Pb remaining.

You are a chemist requested to review and analyse certain debris found on thebottom of aboot found in a crime scene. The debris contains metal fragments. If you identify the metal fragments---it will help the detctive identify the locations traveled by the person who did the crime.

from the bottom of the boot---a sample of unknow metal, X(s) and react it with Oxygen gas according to the following equation: 2x(s) + O2 >>> 2XO(s)

if you react 397.2 grams of X(s) with 100.0 grams of O2(g) to produce 497.2 grams of XO(s), then what is the identity of metal X???

which is it based on the items below

iron
copper
gold
silver
lead

You are a chemist requested to review and analyse certain debris found on thebottom of aboot found in a crime scene. The debris contains metal fragments. If you identify the metal fragments---it will help the detctive identify the locations traveled by the person who did the crime.

from the bottom of the boot---a sample of unknow metal, X(s) and react it with Oxygen gas according to the following equation: 2x(s) + O2 >>> 2XO(s)

if you react 397.2 grams of X(s) with 100.0 grams of O2(g) to produce 497.2 grams of XO(s), then what is the identity of metal X???

which is it based on the items below

iron
copper
gold
silver
lead

........2X + O2 ==> 2XO

.....397.2..100.0....497.2
Note that the grams add to the same on both side; therefore, you know neither X nor O2 was in excess. Thus it is not a limiting reagent problem.
How many moles O2 did you use. That is 100/32 = 3.125
According to the equation, how many mols X were used. That must be 2 x 3.126 = 6.250.
Then since moles = grams/atomic mass we can rearrange to atomic mass = grams/moles.
Substitute and look up the atomic mass on the periodic chart. I vote for Cu.

To determine whether you added enough potassium chloride to remove all the lead(II) ions from the solution, we need to calculate the molarity of lead(II) ions before and after the addition of the potassium chloride solution.

First, let's calculate the number of moles of lead(II) nitrate in the tank:
Molarity of lead(II) nitrate = 2.5 x 10^-6 M
Volume of the tank = 1.5 x 10^6 liters

Number of moles of lead(II) nitrate = Molarity x Volume
= (2.5 x 10^-6 M)(1.5 x 10^6 L)
= 3.75 moles

Next, let's calculate the number of moles of lead(II) ions in 59.7 liters of 0.075 M potassium chloride solution:
Molarity of potassium chloride = 0.075 M
Volume of potassium chloride solution added = 59.7 liters
Number of moles of potassium chloride = Molarity x Volume
= (0.075 M)(59.7 L)
= 4.4785 moles

Since we are interested in the amount of potassium chloride required to remove all the lead(II) ions from the solution, we need to compare the moles of lead(II) nitrate with the moles of potassium chloride.

If the moles of lead(II) nitrate are equal to or greater than the moles of potassium chloride, it means you have added enough potassium chloride to remove all the lead(II) ions from the solution. Conversely, if the moles of lead(II) nitrate are less than the moles of potassium chloride, it means you have not added enough potassium chloride to remove all the lead(II) ions.

Comparing the number of moles:
Moles of lead(II) nitrate = 3.75 moles
Moles of potassium chloride = 4.4785 moles

Since the moles of lead(II) nitrate (3.75 moles) are less than the moles of potassium chloride (4.4785 moles), it indicates that you have added enough potassium chloride to remove all the lead(II) ions from the solution.