Solution Stoichiometry

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!