Given a #0.450*g# mass of lead nitrate, what mass of sodium iodide is required to precipitate the metal as it iodide salt?
I think you mean to ask how much
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.
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: