Title: Stoichiometry Problem
1Stoichiometry Problem
2Sodium phosphate barium nitrate
- Its a double replacement reaction, since we have
2 ionic compounds. - Try to come up with the formulas before you go to
the next slide
3Na3PO4 Ba(NO3)2
- Those are the reactants, now try and predict the
products before going to the next slide. - Remember AX BY ? AY BX
4Na3PO4 Ba(NO3)2 ? Ba3(PO4)2 NaNO3
- Ok, there is the reaction.
- Now before we can do any kind of stoichiometric
problem, we need a balanced chemical equation. - Is it balanced?
52 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Now, its balanced.
- OK, here is your problem
- 3.50 g of sodium phosphate are reacted with 6.40
g of barium nitrate. What mass of barium
phosphate can be produced?
62 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Things to remember
- 1. Everything must be calculated in moles.
- 2. They gave us mass therefore we must convert
to moles. - 3. To figure out moles of sodium phosphate and
barium nitrate, I need to know the molar mass of
each. - Calculate those now, before you go to the next
slide.
72 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
82 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Na3PO4 Molar mass 164 g/mole
- Ba(NO3)2 Molar mass 261.3 g/mole
- Now find moles of each
92 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Na3PO4 3.50 g / 164 g/mole
- We have 0.021 moles of sodium phosphate.
- Ba(NO3)2 6.40 g / 261.3 g/mole
- We have 0.024 moles of barium nitrate.
102 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Now what?
- Well, now you look at your equation.
- For every 2 moles of sodium phosphate, we need 3
moles of barium nitrate to react with it.
Correct?
112 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- We have 0.021 moles of sodium phosphate and 0.024
moles of barium nitrate. Which one are we going
to run out of first? - Think about it before you go to the next slide.
122 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Yes, of course, were going to run out of barium
nitrate first. - To calculate that, lets look at moles of sodium
phosphate and figure out how much barium nitrate
we need to react with all of it.
132 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- 0.021 moles Na3PO4 x (3 moles Ba(NO3)2 / 2 moles
Na3PO4 ) 0.032 moles barium nitrate. - But we dont have that much barium nitrate.
Therefore, were going to run out of barium
nitrate and have some sodium phosphate left over,
after the reaction is done. So, barium nitrate
is the LIMITING REAGENT.
142 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- Using the moles of barium nitrate, we next need
to calculate how much barium phosphate we can
produce. - 0.024 moles Ba(NO3)2 x (1 mole Ba3(PO4)2 / 3
molesBa(NO3)2 ) 0.008 moles barium phosphate.
152 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- The equation tells me that for every 3 moles of
barium nitrate you will produce 1 mole of barium
phosphate. Thats why I picked the factor I did. - So, I will produce 0.008 moles of barium
phosphate. But the problem asked me for MASS,
not moles.
162 Na3PO4 3 Ba(NO3)2 ? Ba3(PO4)2 6 NaNO3
- 0.008 moles of barium phosphate x the molar mass
of barium phosphate should give me the answer. - Did you get 601.9 g/mole for the molar mass? Did
you get 4.82 g of Ba3(PO4)2 for the answer?
17STOICHIOMETRY STEPS
- Thats how you do it.
- Remember, if they give you grams, convert to
moles. If they give you volume of a gas in
liter, convert to moles. If they give you
particles, convert to moles. THATS STEP 1.
18STOICHIOMETRY STEPS
- Step 2 is to use a balanced chemical equation to
determine the stoichiometric relationship. The
balanced equation tells you how many moles of
each substance participate in the reaction. - USE A FACTOR!! Its easy! And now you have your
answer in moles. If they want moles, you are
done. If not, go to step 3.
19STOICHIOMETRY STEPS
- Step 3!
- If they want the answer in grams, convert to
grams. If they want the answer in liters,
convert to liters. If they want the answer in
particles, convert to particles. THATS CHAPTER
7!! (Remember the triangular chart I gave you?)
20The End
- Now practice, practice, practice and then if you
think you know it, - Practice some more!!