What's the theoretical yield of precipitate of a reaction of 0.006 moles of Pb(NO3)2 and 0.003 moles of NaI?

Question

If it helps, the actual yield is 1.55 cm, and I need the theoretical yield in moles.

I appreciate any help

I got the yield as the height of the precipitate, I'm doing an experiment to compare the actual height to the theoretical yield in moles.

Answer 1

Pb(NO3)2 + 2 NaI ---> PbI2 +2 NO3Na

so you need 2 moles of NaI for obtaining 1 mole of PbI2

As youhave only 0.003 moles of NaI, you obtain 0.003/2 = 0.0015mole of PbI2

Answer 2

I am totally lost by the notion of measuring yield in centimeters -- that makes no sense. In the reaction, the product is PbI2; the limiting reagent is NaI, and two moles of that are required per mole of product, so the amount of lead iodide would be 0.0015 moles.

Answer 3

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Answer 4

First of all, how did you get a yield in cm? Considering that's a unit of length...help us out....

Second, you need a balanced reaction first, and I don't see one. Minus 10....

Third, can you tell us what precipitates, and therefore what you're calculating the yield OF? (I can. Hint: Most lead salts are insoluble...except nitrates for example)

Let's see you set this problem up correctly, then we'll help you use the problem to get to the answer of your question. Most people here don't take well to doing people's homework for them....

Glad to HELP. Won't DO it for you.