Download Stoichiometry - Chemistry 11

Stoichiometry
Stoichiometry is the calculation of quantitative (measurable) relationships of
the reactants and products in a balanced chemical reaction (chemicals). It
can be used to calculate quantities such as the amount of products that can
be produced with the given reactants and percent yield.
Mole Relationships in Chemical Equations
Chemical equations are the shorthand method for describing chemical change. What are
some alternate ways of expressing the following chemical equation?
1 N2
+
3 H2(g)

2 NH3 (g)
The above equation tells us that
1 molecule of N2 reacts with 3 molecules of H2 to give 2 molecules of NH3
You can multiply the ratio by the Avogadro constant to obtain:
1 molecule N2 x NA : 3 molecules of H2 x NA : 2 molecules of NH3 x NA
To give:
1 mol N2: 3 mol of H2: 2 mol NH3
The ratio of moles of molecules is identical to the ratio of molecules - it has to be, since
equal numbers of moles have equal numbers of molecules.
The relationship between moles in a balanced chemical equation are called mole ratios.
You can manipulate mole ratios to determine unknown molar amounts in a chemical
reaction.
Example 1:
2 Al(s) + 3Br2(l) → 2AlBr3(s)
How many moles of Br2 are needed to produce 5.0 mol of AlBr3 in excess Al?
? n Br2 = 5 mol x 3 mol Br2
2 mol AlBr3
= 7.5 mol Br2
Example 2:
How many moles of Al are needed to produce 5.0 mol of AlBr3 in excess Br2?
1
? n Al = 5 mol x 2 mol Al
2 mol AlBr3
= 5.0 mol Al
Practice:
Given the following equation
2C2H6 + 7O2 → 4CO2 + 6 H20
a) How many moles of O2 are needed to react with 13.9 mol of C2H6?
? n O2 = 13.9 mol C2H6 x 7 mol O2
2 mol C2H6
= 48.7 mol
b) How many moles of H20 would be produced by 1.4 mol of O2 in excess ethane?
? n H2O = 1.4 mol O2 x 6 mol H2O
7 mol O2
= 1.2 mol
Mass Relationships in Chemical Equations
The balanced chemical equation tells us the mole ratios of the reactants and products.
Therefore, if you know the amount of one substance (in particles, moles or mass) you can
calculate the amount of any other substance using the information in these mole ratios from the
balanced equation.
EXAMPLE 1: How many grams of carbon dioxide would be produced from 60.0 g of carbon
and excess oxygen?
C(s)
60g
+
O2 (g)
→
CO2 (g)
?g
↓ Molar Mass C
↑ Molar mass O2
mol C
mol O2
→
Mole Ratios
(Note: Round to sig figures at the end of the problem)
Step 1; Find moles of your given species from grams given and the molar mass.
? n C = 60.0 g
12 g/mol
= 5 mol
Step 2: Find moles of your required species from the mole ratio.
? n O2 = 5 mol C x 1 mol CO2
1 mol C
= 5 mol
2
Step 3: Find grams of your required species from the molar mass.
? g CO2 = 5 mol x 44 g/mol
= 220 g
= 2.20 x102 g
Example 2. Octane reacts with oxygen to produce carbon dioxide and oxygen.
2C8H18 + 25O2 → 16CO2 + 18H20
How many grams of carbon dioxide are formed in the burning of 10.0 g of gasoline?
2C8H18 + 25O2 →
10.0 g
16CO2 + 18H20
?g
↓ Molar Mass C8H18
↑ Molar mass CO2
mol C8H18
→
mol CO2
Mole Ratios
? n C8H18 =
10.0 g C8H18
114g/mol C8H18
= 0.087719298
Molar Mass C8H18 = (12x8 + 1x18) = 114
? n CO2 = 0.087719298 mol C8H18 x 16 mol CO2
2 mol C8H18
= 0.701754386
? CO2 g = 0.701754386 mol CO2 x 44 g/mol CO2
= 30.9 g
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