If 82.9 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.179 g of precipitate, what is the molarity of lead(II) ion in the original solution?
The equation for the reaction is
Let's start by calculating the moles of
Now we can use the molar ratio from the equation to calculate the moles of
And now we can calculate the molarity of the