Download 1 mol H 2 O - Davis.k12.ut.us

April 3, 2014
•Stoichiometry
Stoichiometry
• Stoichiometry is the study of quantities of materials
consumed and produced in chemical reactions
• Stoikheion (Greek, “element”) +
• -metry (English, measurement)
Stoichiometry
• Stoichiometry is used to make predictions about how much
product will be formed without actually doing the reaction.
• What you need:
• A balanced equation!!
2 H2 + O2  2 H2O
• 2 molecules of H2 react with 1 molecule of O2
to form 2 molecules of water
IT ALSO MEANS
• 2 moles of H2 react with 1 mole of O2 to form 2
moles of water
Mole Ratios
• Once you have a balanced reaction,
you can use mole ratios to convert • Hydrogen to oxygen: 2 mol H
2
from amount of one substance to
1 mol O2
amount of another substance
• Hydrogen to water:
• Oxygen to water:
• Oxygen to hydrogen:
• Water to oxygen:
2H +O 2H O
2
2
2
Mole Ratios
• Once you have a balanced reaction,
you can use mole ratios to convert • Hydrogen to oxygen: 2 mol H
2
from amount of one substance to
1 mol O2
amount of another substance
• Hydrogen to water: 2 mol H2
2 mol H2O
2 H2 + O2  2 H2O
• Oxygen to water:
• Oxygen to hydrogen:
• Water to oxygen:
Mole Ratios
• Once you have a balanced reaction, • Hydrogen to oxygen: 2 mol H2
you can use mole ratios to convert
1 mol O2
from amount of one substance to
• Hydrogen to water: 2 mol H2
amount of another substance
2 mol H2O
2 H2 + O2  2 H2O
• Oxygen to water: 1 mol O2
2 mol H2O
• Oxygen to hydrogen:
• Water to oxygen:
Mole Ratios
• Once you have a balanced reaction, you • Hydrogen to oxygen: 2 mol H2
can use mole ratios to convert from
1 mol O2
amount of one substance to amount of
• Hydrogen to water: 2 mol H2
another substance
2 mol H2O
2 H2 + O2  2 H2O
• Oxygen to water: 1 mol O2
2 mol H2O
• Oxygen to hydrogen: 1 mol O2
2 mol H2
• Water to oxygen:
Mole Ratios
• Once you have a balanced reaction, you • Hydrogen to oxygen: 2 mol H2
can use mole ratios to convert from
1 mol O2
amount of one substance to amount of
• Hydrogen to water: 2 mol H2
another substance
2 mol H2O
2 H2 + O2  2 H2O
• Oxygen to water: 1 mol O2
2 mol H2O
• Oxygen to hydrogen: 1 mol O2
2 mol H2
• Water to oxygen: 2 mol H2O
1 mol O2
Calculating Masses of Reactants and
Products
1.
2.
3.
4.
5.
Make sure the equation is balanced.
Identify your given and your unknown.
Convert mass to moles, if necessary.
Set up mole ratios.
Use mole ratios to calculate moles of the
product/reactant desired.
6. Convert moles to mass, if necessary.
Important
• The “given” and “unknown” can both be reactants,
both be products, or one reactant and one product
• Reaction stoichiometry will ALWAYS have a mol
(given)  mol (unknown) conversion!!!
• To calculate mol (given)  mol (unknown), use mole
ratio, which comes directly from the coefficients of
the balanced chemical equation
• Other conversion factors may (or may not) be
necessary before or after this conversion
MOL (GIVEN)  MOL (UNKNOWN)
• 1. 2H2 + O2  2H2O
• A. How many moles of water are produced from 1 mole of H2?
Mole ratio H2 to H2O : 2 mol H2 = 2 mol H2O
1 mol H2
| 2 mol H2O = 1 x 2 = 1 mol H2O
2 mol H2
2
MOL (GIVEN)  MOL (UNKNOWN)
• 1. 2H2 + O2  2H2O
• B. How many moles of water are produced from 3 moles of O2?
Mole ratio O2 to H2O : 1 mol O2 = 2 mol H2O
3 mol O2
| 2 mol H2O = 3 x 2 = 6 mol H2O
1 mol O2
1
MOL (GIVEN)  MOL (UNKNOWN)  MASS
(UNKNOWN)
• 2. CH4 + 2O2  CO2 + 2H2O
• A. How many grams of water are produced from one mole of CH4?
(molar mass H2O = 18.02 g/mol)
Mole ratio CH4 to H2O : 1 mol CH4 = 2 mol H2O
1 mol CH4 | 2 mol H2O | 18.02 g = 1 x 2 x 18.02 = 36.04 g H2O
1 mol CH4 1 mol H2O
1x1
MOL (GIVEN)  MOL (UNKNOWN)  MASS
(UNKNOWN)
• 2. CH4 + 2O2  CO2 + 2H2O
• B. How many grams of O2 are required to produce 5 mol CO2? (molar
mass O2 = 32.00 g/mol)
Mole ratio CO2 to O2 : 1 mol CO2 = 2 mol O2
5 mol CO2 | 2 mol O2 | 32.00 g = 5 x 2 x 32.00 = 320 g O2
1 mol CO2 1 mol O2
1x1
mass, g (given)  mol (given)  mol
(unknown)
• 3. CH4 + 2O2  CO2 + 2H2O
• A. How many moles of CO2 can be produced from 4.0 g CH4? (molar
mass CH4 = 16.05 g/mol)
Mole ratio CH4 to CO2 : 1 mol CH4 = 1 mol CO2
4.0 g CH4 | 1 mol CH4 | 1 mol CO2 = 4.0 x 1 x 1 = 0.25 mol CO2
16.05 g CH4 1 mol CH4 16.05 x 1
mass, g (given)  mol (given)  mol
(unknown)
• 3. CH4 + 2O2  CO2 + 2H2O
• B. How many moles of CH4 are required to react with 2.0 g O2? (molar
mass O2 = 32.00 g/mol)
Mole ratio O2 to CH4: 2 mol O2 = 1 mol CH4
2.0 g O2 | 1 mol O2 | 1 mol CH4 = 2.0 x 1 x 1 = 0.031 mol CH4
32.00 g O2 2 mol O2
32.00 x 2
mass, g (given)  mol (given)  mol
(unknown)  mass, g (unknown)
• 4.
____NH3 + ____O2  ____NO + ____H2O
First you must balance the equation.
N: 1
=
N: 1
H: 3
≠
H: 2
O: 2
=
O: 2
mass, g (given)  mol (given)  mol
(unknown)  mass, g (unknown)
• 4.
__2__NH3 + ____O2  ____NO + __3__H2O
N: 1 2
H: 3 6
O: 2
≠
≠
≠
N: 1
H: 2 6
O: 3 4
mass, g (given)  mol (given)  mol
(unknown)  mass, g (unknown)
• 4.
__2__NH3 + ____O2  _2__NO + __3__H2O
N: 1 2
H: 3 6
O: 2
=
=
≠
N: 1 2
H: 2 6
O: 3 4 5
Leave oxygen for last!
mass, g (given)  mol (given)  mol
(unknown)  mass, g (unknown)
• 4.
__4__NH3 + __5__O2  _4__NO + __6__H2O
N: 1 2 4
=
N: 1 2 4
H: 3 6 12
=
H: 2 6 12
O: 2 10
≠
O: 3 4 5 10
There is no way to fit 2 evenly into 5. Try doubling everything and then
see if you can make oxygen balance.
mass, g (given)  mol (given)  mol
(unknown)  mass, g (unknown)
• 4. 4 NH3 + 5 O2  4 NO + 6 H2O
• A. How many grams of H2O can be formed from 1.0 g NH3?
(molar mass NH3 = 17.04 g/mol, molar mass H2O = 18.02 g/mol)
Mole ratio NH3 to H2O: 4 mol NH3 = 6 mol H2O
1.0 g NH3 | 1 mol NH3 | 6 mol H2O | 18.02 g = 1.0 x 1 x 6 x 18.02 = 1.59 g H2O
17.04 g NH3 4 mol NH3 1 mol H2O
17.04 x 4
mass, g (given)  mol (given)  mol
(unknown)  mass, g (unknown)
• 4. 4 NH3 + 5 O2  4 NO + 6 H2O
• B. If 5.0 g NO are formed, how many grams of H2O are also formed? (molar
mass NO = 30.01 g/mol)
Mole ratio NO to H2O: 4 mol NO = 6 mol H2O
5.0 g NO| 1 mol NO | 6 mol H2O | 18.02 g = 5.0 x 1 x 6 x 18.02 = 4.50 g H2O
30.01 g NO 4 mol NO 1 mol H2O
30.01 x 4