Transcript Unit 6 Stoichiometry

Stoichiometry – the calculation of
quantities in chemical equations.
When balancing a chemical equation, change
only the coefficients, never change the
subscripts.
The coefficients tell how many molecules
are in a reaction, the subscripts tell how
many atoms are in a molecule.
The coefficients are very important to
stoichiometry-they tell the relationship
between reactants and products in an
equation.
Things that are conserved in every reaction:
•Energy
•Mass
•Number of atoms
The coefficients can be:
•A literal interpretation of how many actual
molecules in an equation.
2 H2 + O2  2 H2O
This means 2 molecules of hydrogen will
react with 1 molecule of oxygen to give 2
molecules of water.
This is good for balancing equations, but
little else as molecules are too small to deal
with in the lab.
•The coefficients can tell the number of
moles of each molecule in the equation.
2 H2 + O2  2 H2O
This means 2 moles of hydrogen reacts with
1 moles of oxygen will give 2 moles of water.
•The coefficients can tell the number of
volumes of gas in a reaction. This will work
only if the molecules are gas molecules.
2 volumes of hydrogen reacts with 1 volume
of oxygen will give 2 volumes of water.
Moles Ratios – a fraction, in moles, of 2
different molecules in a balanced equation (a
conversion factor for unit analysis).
2 H2 + O2  2 H2O
There are three equivalencies (comparisons):
2 mol H2 ≈ 1 mol O2
2 mol H2 ≈ 2 mol H2O
1 mol O2 ≈ 2 mol H2O
From each equivalency, two mole ratios can
be formed:
2 mol H2/1 mol O2
1 mol O2/2 mol H2
2 mol H2/2 mol H2O
2 mol H2O/2 mol H2
1 mol O2/2 mol H2O
2 mol H2O/1 mol O2
Which mol ratio is used depends on which
ratio is needed when doing unit analysis, or
which one the question asks for.
N2 + H2  NH3
If you start with 3.0 moles of N2, how many
moles of H2 are needed? How many moles
of NH3 can be produced?
1. Balance the reaction:
N2 + 3 H2  2 NH3
2. Write out the equivalencies:
1 mol N2≈3 mol H2; 1 mol N2≈2 mol NH3;
3 mol H2≈2 NH3
3. Set up the unit analysis
3.0 mol N2
3 mol H2 = 9.0 mol H2
1 mol N2
3.0 mol N2
2 mol NH3 = 6.0 mol NH3
1 mol N2
The mole ratios are conversion factors using
the coefficients from the balanced equation.
The Double Flux Capacitor
Part. A
Mass A
Part. B
Mass A
Mole A
Moles B
Vol. A
Vol. B
Particles  Moles
Mass 
Moles
Volume  Moles
Moles A  Moles B
1 Mol = 6.022 E 23 things
1 Mol = Molar Mass
1 Mol = 22.4 L gas
coefficients from the balanced
equation (mole ratios)
What mass of I2 can be made from 10.0
liters of Cl2(g) according to the following
reaction: Cl2 + 2 KI  2 KCl + I2
10.0 L Cl2
1 mol Cl2
1 mol I2
253.8 g I2 = 113 g I2
22.4 L Cl2
1 mol Cl2
1 mol I2
1. How many grams of water can be made from 13.5 grams
of H3PO4 according to the following reaction:
H3PO4  H4P2O7 + H2O
2. What volume of hydrogen gas is needed to make 15.3
grams of iron according to the reaction:
Fe2O3 + H2  Fe + H2O
3. How many liters of CO2 can be made from 17.5 grams of
oxygen:
C20H42 + O2  CO2 + H2O
4. How many molecules of Fe2O3 are needed to react with
18.5 liters of CO according to the reaction:
Fe2O3 + CO  Fe + CO2
5. How many grams of PbI4 can be made when 12.5 g CrI3
react:
PbO2 + CrI3  PbI4 + Cr2O3
Theoretical Yield – the maximum amount of
product that can be formed based upon a
balanced equation (using the double flux
capacitor).
Actual Yield – the amount of product
actually made in the lab.
% Yield – (actual yield/theo. yield) x 100 %
The % yield should always be less than 100
% due to experimental error. When plugging
in the theoretical yield, always use the
UNROUNDED value.
The actual yield will always be one of the
products. It can never be one of the
reactants.
Na + H2O  NaOH + H2
If 18.3 g of sodium can produce 8.23 liters
of hydrogen gas, what is the % yield?
Balanced equation: 2,2,2,1
18.3 g Na
1 mol Na
1 mol H2
22.4 l H2 = 8.91 l H2
23.0 g Na
2 mol Na
1 mol H2
(8.23 l H2 / 8.9113 … l H2) x 100 % = 92.4 %
Limiting Reagent (Reactant) – the starting
material that runs out first in a chemical
reaction causing the reaction to stop.
2 B + 1 C + 3 H + 1 T + 3 L + 1 M  1 HS
How many ham sandwiches can be made from
10 bread, 8 cheese, 12 ham, 7 tomato, 15
lettuce and 20 mustard?
Bread  5
cheese  8
ham  4
Tomato  7
lettuce  5
mustard  20
Ham is the limiting reactant (LR) because
only 4 ham sandwiches can be made. After
that point, the ham runs out causing the
reaction to stop. Therefore no more
sandwiches can be made.
When working a limiting reactant problem,
work the problem twice- calculate the
amount of product that can be made from
each reactant. The smallest amount of
product will be how much that can be made.
The reactant that causes the least amount
of product will be the LR.
N2 + 3 H2  2 NH3
What mass of NH3 can be made from 6.37 g
of N2 and 10.3 g of H2? What is the LR?
6.37 g N2
10.3 g H2
1 mol N2 2 mol NH3
17.0 g NH3 = 7.74 g NH3
28.0 g N2 1 mol N2
1 mol NH3
1 mol H2
2 mol NH3 17.0 g NH3 = 58 g NH3
2.0 g H2 3 mol H2
1 mol NH3
7.74 grams of NH3 can be made in the reaction, and N2 is
the LR.
Stoichiometry
• Only 1 given number – double flux
capacitor problem. “theoretical yield”
• 2 given numbers: 1 reactant and 1 productdouble flux capacitor with the reactant
going to same units as the product (theo
yield). Product number is the actual yield.
(actual/theo) x 100 % = % yield
• 2 given numbers, both reactants. Limiting
reagent problem. Double flux each
starting number. Smallest product is the
answer. Starting number that gives
smallest product is LR.