# Given a #0.450*g# mass of lead nitrate, what mass of sodium iodide is required to precipitate the metal as it iodide salt?

##### 2 Answers

#### Answer:

I think you mean to ask how much

#### Explanation:

With respect to

And thus, for metathesis, we need a TWICE molar quantity of

Anyway, if I have misinterpreted your problem, send me a beatdown.

#### Answer:

#### Explanation:

Given:

We begin the factor-label method by writing the given over 1:

We look up the molar mass of

Please observe how the units cancel:

From the balanced equation, we observe that 2 moles of

Again, please observe how we have cancelled the units as we progress toward the conversion:

Finally, we look up the molar mass of

With a final cancellation of units, please observe that the only units that remain are grams of Potassium Iodide:

We merely perform the multiplications and divisions to obtain the answer: