Molarity and Stoichiometry

So, today we've taken stoichiometry one step further, which is including the molarity.

Remember the stuff we did last day on stoichiometry was basically to take gram stuff into the mole stuff and such and such, but today, we involved molarity.

As we know, molarity is a concentration stuff, so for example:

-How many mL of 0.124M NaOH contain enough NaOH to react with 15.4 mL of 0.108 M H2SO4 ?

-First of all, we have to wrtie the chemical reaction out and balance it.

    #mL       15.4mL
2 NaOH + H2SO4 ---> 2 H2O + Na2SO4
0.124 mol   0.108 mol

--------------- --------------

     1L               1L

and the question has given out a lot of information, as you can see, it is asking the milliliter of NaOH (in red). And we all know the molarity can be written as moles per L, so the question said 0.124 M, which is equivelent to 0.124 mole per L (in yellow). and another piece of information is 15.4 mL (in green) of 0.108M, or 0.108 mole per L (in aqua).

And now let's start doing some conversion!!


 15.4 mL           1 L          0.108 mol H2SO4     2 mol NaOH     
-------------- x --------------- x ----------------------------- x ---------------------
   1 L           1000 mL                   1 L              1 mol H2SO4   
                  1 L             1000mL
    x ------------------------- x ------------ = 26.82580645 mL NaOH   <---- WRONG!
       0.124 mol NaOH        1 L

Be careful !! sig fig counts for mark!!

26.82580645 mL NaOH, 3 sig figs = 26.8 mL NaOH

+

 

YAY !