[Mikiricounter]

2NaI + Pb(NO3)2     PbI2 + 2NaNO3

If 4.61 grams of PbI2 was isolated from the reaction of 2NaI and Pb(NO3)2, what will be the calculated amount of NaI that reacted with PbI2?

Molar Mass:
PbI2 = 330 g/mol
NaI = 150 g/mol

The answer is 3.00 grams.

Can anyone please tell me the step by step solution to this problem? This question was given as a review pointer to us but there was only an answer but no solution. It would be really helpful to know how things get done in this problem. Thank you very much!

[AWK]

Calculate moles of PbI₂

[Mikiricounter]

Hello! I tried working out the problem based on what you said but the answer I got was something 295.5 g. So this is what I did after the hint you gave me.

First, I subtracted 4.61 g from 330 g/mol (= 325.39 g) then I converted it to moles (0.986 moles of PbI2). Next, I converted the moles of PbI2 that I got (0.986 moles of PbI2) to the moles of NaI and got 1.97 moles of NaI. (I just multiplied it by 2). I then decided to convert it to grams by multiplying 1.97 moles of NaI to 150 g of NaI. I got the answer of 295.5 g.

I did it wrong didn't I? What is the correct way?

[AWK]

Subtraction only sometimes gives the same result as division. It didn't work this time.
The rest of the calculations would be correct if the first number were correct.


Moreover, the problem shows 2 errors
If 4.61 grams of PbI2 was isolated from the reaction of 2NaI and Pb(NO3)2, what will be the calculated amount of NaI that reacted with PbI2(should be Pb(NO₃)₂)?

and

PbI2 = 330 (should be 461) g/mol

With correct number, problem can be solved without calculator!

[Borek]

First, I subtracted 4.61 g from 330 g/mol (= 325.39 g)

What for? What were you trying to do?

[AWK]

You got a small part of mole PbI₂ (207+2*127)
4.61/461 = ..x.. This amount of lead iodide is obtained from potassium iodide (2*x moles), which multiplied by the mass of KI = 150 g/mol will give you a correct answer.