UNIT 6: Stoichiometry
Review
Molar Mass Review
• Example:
11/28/2011
11/28/2011
UNIT 6: Stoichiometry
Mole to Mole
Mole to Mole
• First write the balanced chemical equation.
• Second use the coefficients in the balanced chemical equation to write the mole to mole ratio.
• Example:
• If we have 5.0 moles Al2O3 decomposing, how many moles of O2 will we have?
• 2Al2O3 à 4Al +
3O2
5.0 mol Al2O3 3 mol O2
2 mol Al2O3
= 7.5 mol O2
11/29/2011
UNIT 6: Stoichiometry
Mole to Mass
Mole to Mass
• In mole­mass calculations, you calculated the number of moles and then convert it to grams using the molar mass.
• Example:
• 13.0 moles of Al2O3 decompose to form how many grams of O2 gas?
Molar mass of O • 2Al2O3 à 4Al + 3O2
(2 * 16.00g)
2
13.0 mol Al2O3
3 mol O2
32.00 g O2
2 mol Al2O3
1 mol O2
= 624 g O2
11/30/2011
UNIT 6: Stoichiometry
Mass to Mole
Mass to Mole
• In mass­mole calculations, you first convert grams into moles using molar mass, and then calculate the number of moles that reacted or were produced.
• Example:
• If 2.00 grams of Al are produced, how many moles of Al2O3 did we start with?
• 2Al2O3 à 4Al + 3O2
2.00 g Al
1 mol Al
2 mol Al2O3
26.98 g Al
4 mol Al
= 0.0371 mol Al2O3
12/1/2011
UNIT 6: Stoichiometry
Mass to Mass
Mass to Mass • In mass­mass calculations, you need to convert your given mass to moles, and then determine your unknown in moles. Then you need to convert the number of moles you solved for into grams.
• Example:
• If 3.00 grams of Al2O3 decompose, how many grams of Al are produced?
• 2Al2O3 à 4Al +
3.00 g Al2O3
1 mol Al2O3
3O2
4 mol Al
101.96 g Al2O3 2 mol Al2O3
26.98 g Al
1 mol Al
=
1.59 g Al
12/5/2011
UNIT 6: Stoichiometry
Mixed Stoichiometry
Mixed Stoichiometry
(Mass)
1 mol of A
g of A
(Moles)
mol of B
mol of A
(Mass)
• Example:
g of B
1 mol of B
(Moles)
UNIT 6: Stoichiometry
Mixed Stoichiometry
Mixed Stoichiometry
• Example:
12/5/2011
12/7/2011
UNIT 6: Stoichiometry
Limiting Reactant
Limiting Reactant
• Limiting reactant = the reactant that will be used up first in the reaction, it limits the amount of product that can be produced • Excess reactant = the reactant that will be left over after the reaction, it does not affect the amount of product that can be produced
• Example:
• 45.0 grams of water reacts with 44.0 grams of carbon dioxide. Which compound limits the reaction? How many grams of hydrogen carbonate are produced?
• H2O
+
CO2 à H2CO3
12/7/2011
UNIT 6: Stoichiometry
Limiting Reactants
Limiting Reactants
(continued)
• Example: (continued)
45.0 g H2O 1 mol H2O 1 mol H2CO3
18.016 g H2O 1 mol H2O
44.0 g CO2 1 mol CO2 1 mol H2CO3
44.01 g CO2 1 mol CO2
1.00 mol H2CO3 62.03 g H2CO3
1 mol H2CO3
= 2.50 mol H CO
= 1.00 mol H2CO3
2
3
= 62.0 g H2CO3
UNIT 6: Stoichiometry
Limiting Reactants
Limiting Reactants
(continued)
• Example: 12/8/2011
12/12/2011
UNIT 6: Stoichiometry
Percent Yield
Percent Yield
• Percent Yield = a mathematical comparison between the experimental amount (actual) of product and the calculated (theoretical) amount.
• % Y = actual results x 100%
theoretical results
• Example:
• If 34.5 grams of sodium are reacted with an excess of chlorine and 43.9 grams of salt are actually produced, what is the percent yield?
• 2Na + Cl2 ­­­> 2NaCl
34.5 g Na
1 mol Na
22.99 g Na
2 mol NaCl 58.44 g NaCl
2 mol Na 1 mol NaCl
= 87.8 g NaCl
%Y =
43.9 g NaCl
x 100
87.8 g NaCl
= 50.0%
UNIT 6: Stoichiometry
Percent Yield
Percent Yield
• Example:
12/13/2011