Download Chem 11 Stoichiometry (mol-mol) Using the formulas we have

Chem 11
Stoichiometry (mol-mol)
Using the formulas we have discussed, we can now get into the real stoichiometry
calculations. We will need to look at writing equations and balancing them in order to
properly and correctly carry out all stoichiometric calculations.
Write the reaction for the formation of Ammonia:
N2 + 3H2 2NH3
Proof of the Conservation of Mass:
From the balanced equation, we can say that 1 molecule of N2 and 3 molecules of H2
combine to form 2 molecules of NH3 OR we can say that 1 mole of N2 and 3 moles of
H2 combine to form 2 moles of NH3.
Using the coefficients as moles, we can determine the mass of the each of the reactants
present and also for the product formed.
Mass of Reactants:
m = nxM = (1 mol N2)(28.0134g/mol N2) = 28.0134g N2
m = nxM = (3 mol H2)(2.01588g/mol H2) = 6.04764g H2
Total mass of Reactants = 28.0134g N2 + 6.04764g H2 = 34.06104g
Mass of Products:
m = nxM = (2 mol NH3)(17.03052g/mol NH3) = 34.06104g
Note that the mass of both reactants and products are the same value – Law of
Conservation of Mass!
Balanced equations allow us to calculate the quantities of reactants and products in a
reaction. When you know the quantity (usually grams or moles) of one substance, you
can calculate the quantity of any other substance.
Recall: STOICHIOMETRY is the calculation of quantities in chemical equations.
Consider the information we can get from the following balanced equation:
N2(g)
+
1 molecule
10 molecules
1 x (6.02 x 1023 molecules)
1 mol N2
3H2(g)
3 molecules
30 molecules
3 x (6.02 x 1023 molecules)
3 mol H2
→
2NH3(g)
2 molecules
20 molecules
2 x (6.02 x 1023 molecules)
2 mol NH3
The most important interpretation of this equation is that 1 mole of nitrogen reacts
with 3 moles of hydrogen to form 2 moles of ammonia (as we used to determine the mass
of reactants and products above).
With this information we can relate moles of reactants to moles of products. The
coefficients from the balanced equation can be used to write conversion factors. Then the
number of moles of a product can be calculated from a given number of moles of
reactant. Using other mole-quantity relationships, we can introduce mass, volume, and
particles into our calculations (this is what we will be getting to in the next few classes).
Example 1:
How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with
hydrogen?
Solution 1:
We will work from a known number of moles of N2 to the unknown number of moles of
NH3. From the balanced equation just given, we see that 1 mol of N2 produces 2 mol of
NH3. This relationship can be written and used as a conversion factor.
The conversion factor is called a mole ratio.
1 mol N2_
_2 mol NH3_
OR
2 mol NH3
1 mol N2
To solve the problem, we start with what we’re given, 0.60 mol of N2. Then we multiply
it by the form of the mole ratio that allows the units we want to get rid of to cancel.
_2 mol NH3_
= 1.2 mol NH3
1 mol N2
(Note that mole ratios from balanced equations are considered to be exact. They do not
enter into the determination of significant figures in the answer.)
0.60 mol N2 x
Example 2:
a) How many moles of aluminum are needed to form 2.3 moles of aluminum oxide?
b) Calculate the number of moles of aluminum oxide formed when 17.2 moles of oxygen
reacts with aluminum.
Solution 2:
First we need to write the balanced chemical equation for this reaction:
4 Al + 3 O2 2 Al2O3
Then we use the information given in the question, and our coefficients from our
balanced equation, to solve for the moles of the substance we are asked to find.
a) nAl = (2.3mol Al2O3)(4 mol Al/2 mol Al2O3) = 4.6mol Al
b) n Al2O3 = (17.2mol O2)(2mol Al2O3/3mol O2) = 11.4666mol Al2O3 = 11.5mol Al2O3
Example 3:
a) If 3.84 moles of C2H2 burns completely, how many moles of oxygen are needed?
b) How many moles of carbon dioxide are produced when 2.47 moles of C2H2 are
burned?
Solution 3:
Again, we write the balanced chemical equation first:
2 C2H2 + 5 O2 4 CO2 + 2 H2O
Use our coefficients and the information given in the question to convert to the amount of
moles of the substance as asked in the question.
a) n O2 = (3.84mol C2H2)(5mol O2/2mol C2H2) = 9.60mol O2
b) n CO2 = (2.47mol C2H2)(4mol CO2/2mol C2H2) = 4.94mol CO2
Complete worksheet – Mole to Mole Practice Problems (#’s 1-6)