Calculations (1 Viewer)

[z600]

I have no idea how to do the following and would someone be kind enough to show me

100mL 2M sodium iodine with excess of 1M lead (II) nitrate. calculate the mass of precipitate formed by mixing them together.

I know its the easiest thing in the world but i was away and completely missed out on the lessons

thanks!

 

[Trebla]

2NaI + Pb(NO₃)₂ --> PbI₂ + 2NaNO₃

The precipitate is the lead (II) iodide (since all nitrates are soluble)

First convert everything into moles:
n(NaI) = 0.1 x 2 = 0.2 moles

Since lead nitrate is in excess, the limiting agent is sodium iodide

By stoichiometric ratios, the number of moles of lead iodide is 0.1. So convert the moles into mass so you get: 0.1 x (207.2 + 2 x 126.9) = 46.1 grams

I think that's right....someone please check for me....