Solution Stoich
By Megan Noone
Date: November 8, 2012
Homework: Solution Stoichiometry Sheet, (It's double-sided!)
Next Test: Tuesday November 13, 2012.
Today we learned how to apply Stoichiometry to Molarity!
For example...
What volume of .1 M Na3PO4 is required to precipitate all of the lead (II) ions from 150 ml of .250 M Pb(NO3)2?
We can solve this problem by using what we learned in Stoichiometry and combining it with what we learned from Molarity!
1) Balance the equation!
2 Na3PO4 + 3 Pb(NO3)2 ------> 6 NaNO3 (aq) + Pb3(PO4)2 (s)
2) Convert the volume to moles!
150ml x 1L x 0.250 mol Pb(NO3)2
1000ml 1L
3) Next convert the moles of Pb(NO3)2 to moles of Na3PO4!
150ml x 1L x 0.250 mol Pb(NO3)2 x 2 mol Na3PO4
1000ml 1L 3 mol Pb(NO3)2
4) Finally, convert moles of Na3PO4 to liters!
150ml x 1L x 0.250 mol Pb(NO3)2 x 2 mol Na3PO4 x 1L
1000ml 1L 3 mol Pb(NO3)2 0.1 mol Na3PO4
When all is calculated out, it looks something like this!
150ml x 1L x 0.250 mol Pb(NO3)2 x 2 mol Na3PO4 x 1L = 0.25 L Na3PO4
1000ml 1L 3 mol Pb(NO3)2 0.1 mol Na3PO4
See? That wasn't so hard!
As a reward, please enjoy this picture of 2 adorable penguins!
Good luck with the homework!
