The Math of Equations Stoichiometry the calculation of
quantities in chemical reactions
Slide 3
Limiting Reactant 1. Calculate the amount of product that could
be made for each reactant 2. Compare the amount that could be made
to the amount of product actually made to determine which reactant
is limiting 3. Calculate the excess reactant by calculating the
amount that actually reacts and then subtracting
Slide 4
Limiting Reactant There is always some unreacted chemical
Because you cannot measure to the last atom So one reactant gets
used up first Once one reactant is used up, no more product can be
made The Chemical used up first is the LIMITING REACTANT
Slide 5
So.. reactant that is not all used up in the reaction Excess
reactant
Slide 6
1.Determine the limiting reactant when.. 4 grams of H 2 are
reacted with 31g of O 2. Determine moles of products each reactant
can form The one that is limiting will form the least amount of
product 2H 2 +O 2 2H 2 O 4g H 2 x 1 mole H 2 x 2 moles H 2 O = 2
moles H 2 O 2 g H 2 2 moles H 2 31g O 2 x 1 mole O 2 x 2 moles H 2
O = 1.93 moles H 2 O 32 g O 2 1 moles O 2 Oxygen would run out
first, thats why it makes less water. Oxygen is the limiting
reactant
Slide 7
Mg + 2HCl MgCl 2 + H 2 3.500 g of Mg is added to 300.0 mL of
1.000 M HCl How many grams of Magnesium chloride are formed? 3.500g
Mg x 1 mole x 1 mole MgCl 2 x 95.21 g MgCl 2 = 13.71g MgCl 2 24.31
1 mole Mg mole MgCl 2 0.3000L x 1.000mole x 1 mole MgCl 2 x 95.21 g
MgCl 2 = 14.28gMgCl 2 1 L 2 mole HCl mole MgCl 2 Since Mg runs out
first, it is the limiting reactant. What test could you run to
determine there was acid left over?
Slide 8
Mg + 2HCl MgCl 2 + H 2 3.5 g of Mg is added to 300.0 mL of 1.0M
HCl How many liters of gas @ STP? Since magnesium is the limiting
reactant 3.500g Mg x 1 mole x 1 mole H 2 x 22.4 L H 2 = 6.450 L H 2
24.31 1 mole Mg mole H 2
Slide 9
Mg + 2HCl MgCl 2 + H 2 What is left over and how many moles? If
magnesium is the LR, then HCl is left over 3.500g Mg x 1 mole x 2
mole HCl =.2879 moles HCl used in reaction 24.31 1 mole Mg 0.3000L
x 1.000mole HCl =.3000 moles HCl given 1 L 0.3000 moles HCl -
0.2879 (moles HCl used by Mg) =.0121 mols HCl 0.0121moles HCl
=.0403 M.0121moles HCl x 36.5g = 0.4417g 0.3000L mole HCl
Steps in stoichiometry Write symbols and balance Write the
moles on top/known on bottom Circle the known and unknown Do
Dimensional analysis problem
Slide 12
Mole Mole problems Ammonium nitrate decomposes to yield
dinitrogen monoxide and water How many moles of water are produced
if 14.8 moles of ammonium nitrate react? 1. Write symbols and
balance 2. Write the moles on top/known on bottom 3. Circle the
known and unknown 4. Do Dimensional analysis problem
Slide 13
practice Bubbles of hydrogen gas and iron(III) chloride are
produced when iron is dropped into hydrochloric acid How many moles
of hydrogen are produced when 14.5 moles of iron react?
Slide 14
Mole Mass problems Aluminum and silver nitrate react. How many
moles of aluminum nitrate will be produced if 34.6 g of aluminum
react? 1. Write symbols and balance 2. Write the moles on top/known
on bottom 3. Circle the known and unknown 4. Do dimensional
analysis problem
Slide 15
Mass-Mass Lithium nitride reacts with water to produce ammonia
and lithium hydroxide. How many g of LiOH are produced if.38g of Li
3 N react?
Slide 16
Other Sodium chloride is produced when chlorine reacts with
sodium 1.How many L of Chlorine are needed to make 45.6 g sodium
chloride? 2.How many F.U. of NaCl are produced when 1.32 x 10 15
atoms of Na react?
Slide 17
Law of Conservation of Matter Prove the law of conservation of
matter for 2NH 3 N 2 + 3H 2 Do Dimensional analysis problem for
each reactant and product to determine the mass in g
Slide 18
% Yield (product) % Yield = Actual (experiment) x 100%
(product) Theoretical (calculation)
Slide 19
You Try 13.5 g of sodium metal reacts with excess hydrochloric
acid. How many liters of hydrogen are produced (at STP 22.4L / mol)
13.5g Na x 1 mol Na x 1 mol H 2 x 22.4L H 2 = 23.0 g Na 2 mol Na 1
mol H 2 2Na + 2HCl 2 NaCl + H 2 Steps to solve 1.Predict products
2.Balance equation 3.Set up Dimensional analysis 4.Solve 6.57
liters of H 2 gas
Slide 20
4Al + 3O 2 2Al 2 O 3 x 102.g Al 2 O 3 = 1 mol Al 2 O 3 1 mole
27.01 g x 2 mol Al 2 O 3 4 mole Al 28.3 g aluminum oxide 15.0 g of
aluminum are reacted with excess oxygen. How much oxide was
produced? 15.0g x
Slide 21
Matter is Conserved 4g + 32g = 36g If you react 8 grams of
hydrogen and 64 g of oxygen, how much water would did you make?
Four Moles of H 2 O! Or 4 x 18 g/mole = 72 grams 2H 2 + O 2 > 2H
2 O 8g + 64g = ??g 8g + 64g = 72g
Slide 22
How many grams of water are produced from 2 moles of hydrogen
gas and 1 mole of oxygen gas? 2 moles of H 2 = how many grams? 2
mole x 2 atoms x 1 g/mole = 4g 1 mole of O 2 = how many grams? + 2
atom x 16 g/mole = 32 g 2 mole of H 2 O =how many grams 2 mole x 18
g/mole=36g 2H 2 + O 2 > 2H 2 O 4g 32g 36g
