Download Ch. 3 Sections 3.9-3.10 Notes

AP Chemistry
Ch. 3 Sections 3.9-3.10
Stoichiometric Calculations: Amounts of Products and Reactants, Calculations Involving a
Limiting Reactant
Stoichiometric Calculations
• The coefficients in chemical equations represent numbers of molecules or formula units, not
masses of molecules or formula units.
• When a reaction is to be run in a laboratory or chemical plant, the amount of substances
needed cannot be determined by counting molecules directly.
• Counting is always done by weighing.
• We will see how chemical equations can be used to determine the masses of reacting
chemicals.
• 3NaOH (aq) + H3PO4 (aq) → Na3PO4 (aq) + H2O
• In the above reaction, the coefficients give us the ratios by moles of the reactants and
products.
• The coefficients tell us that to make 1 mol of Na3PO4 from 1 mol of H3PO4, we must also use
3 mol of NaOH.
• We don’t however, have to carry out the reaction with these actual numbers of moles.
• Whatever quantities we choose must be in the proportions set by the coefficients.
• Regardless of the scale of the reaction, the coefficients of a chemical equation give the ratio
in which the moles of one substance react with or produce moles of another.
• We can look at this equation as a calculating tool, because its coefficients give us
stoichiometric equivalencies between the substances involved.
• For example, from the above equation we can generate the following equivalencies:
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3 mol NaOH ↔ 1 mol H3PO4
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3 mol NaOH ↔ 1 mol Na3PO4
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3 mol NaOH ↔ 3 mol H2O
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1 mol H3PO4↔ 1 mol Na3PO4
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1 mol H3PO4↔ 3 mol H2O
• Any of these can be used to construct conversion factors called mole ratios for stoichiometric
calculations.
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• Example: How many moles of sodium phosphate, Na3PO4, can be made from 0.240 mol of
NaOH by the following reaction?
3NaOH (aq) + H3PO4 (aq) → Na3PO4 (aq) + H2O
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In practical work, a chemist is often confronted by a question such as the following.
“If I start with so many grams of reactant A, how many grams of reactant B ought I use, and
how many grams of a particular product should be produced?”
Notice that the question concerns grams, not moles, for the practical reason that masses in
grams are delivered by laboratory balances.
The coefficients of the desired reaction, however, know nothing about grams, only about
relative numbers of moles.
If we know two facts, namely, the balanced equation and the mass of any substance in it, we
can calculate the required or expected mass of any other substance in the equation.
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Example: Portland cement is a mixture of the oxides of calcium, aluminum, and silicon.
The raw material for its calcium oxide is calcium carbonate, which occurs as the chief
component of a natural rock, limestone. When calcium carbonate is strongly heated it
decomposes by the following reaction. One product CO2, is driven off to leave the desired
CaO as the only other product.
CaCO3 (s) → CaO (s) + CO2 (g)
A chemistry student is to prepare 1.50 x 102 g of CaO in order to test a particular “recipe” for
portland cement. How many grams of CaCO3 should be used, assuming that all will be
converted?
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Calculating Masses of Reactants and Products in Chemical Reactions
Balance the equation for the reaction.
Convert the known mass of the reactant or product to moles of that substance.
Use the balanced equation to set up the appropriate mole ratios.
Use the appropriate mole ratios to calculate the number of moles of the desired reactant or
product.
Convert from moles back to grams if required by the problem.
Example: One of the most spectacular reactions of aluminum, the thermite reaction, is with
iron oxide, Fe2O3 , by which metallic iron is made. So much heat is generated that the iron
forms in the liquid state. The equation is
Al (s) + Fe2O3 (s) → Al2O3 (s) + Fe (l)
A certain welding operation, used over and over, requires that each time at least 86.0 g of Fe
be produced. What is the minimum mass in grams of Fe2O3 that must be used for each
operation? Calculate also how many grams of aluminum are needed.
Calculations Involving a Limiting Reactant
• Many times a chemist will mix reactants in a mole ratio that does not agree with the
coefficients of the equation.
• Some reactions proceed better when one reactant is in stoichiometric excess, for example.
• One such reaction is the preparation of ammonia, NH3, from its elements.
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N2 (g) + 3H2 (g) → 2NH3 (g)
Suppose a chemist mixed 1.00 mol of N2 with 5.00 mol of H2. What is the maximum
number of moles of product that could form?
Note the coefficients tell us that 1 mol of N2 consumes 3 mol of H2.
1 mol N2 ↔ 3 mol H2
But 5 mol of H2 was used, not 3, so there will be 2 mol of H2 left over.
Once the 1 mol of N2 taken is consumed, no additional NH3 can form.
Therefore, the reactant that is completely consumed limits the amount of product that forms,
so it is called the limiting reactant.
In this reaction, N2 is the limiting reactant, it limited the amount of NH3 that was formed.
Example: In an industrial process for making nitric acid, the first step is the reaction of
ammonia with oxygen at high temperature in the presence of a platinum gauze. Nitrogen
monoxide forms as follows:
4NH3 + 5O2 → 4NO + 6H2O
How many grams of nitrogen monoxide can form if a mixture of 30.00 g of NH3 and 40.00 g
of O2 is taken initially? How many grams of the excess reactant is left over?
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In most experiments designed for chemical synthesis, the amount of a product obtained falls
short of the calculated maximum amount.
Losses occur for several reasons.
Some are mechanical, such as materials sticking to glassware.
But one of the most common causes of obtaining less than the stoichiometric amount of a
product is the occurrence of a competing reaction.
It produces a by product, a substance made by a reaction that competes with the main
reaction.
The actual yield of desired product is simply how much is obtained or isolated, stated in
mass units.
The theoretical yield is the amount of product formed when the limiting reactant is
completely consumed and no losses occur.
When less than the theoretical yield of product is obtained, chemists generally calculate the
percentage yield or percent yield to describe how well the reaction went.
The percent yield is the actual yield calculated as a percentage of the theoretical yield.
Percent yield =
actual yield
x 100%
theoretical yield
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Example: A chemist set up a synthesis of phosphorus trichloride by mixing 12.0 g P with
35.0 g Cl2 and obtained 42.4 g of PCl3. Calculate the percent yield of this compound. The
equation for the main reaction is:
2P (s) + 3Cl2 (g) → 2PCl3 (l)