Transcript Chapter 4 Reaction Stoichiometry • Multiplying the chemical formulas in a balanced • • chemical equation reflect the fact that atoms are neither created nor.

Chapter 4
Reaction Stoichiometry
• Multiplying the chemical formulas in a balanced
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chemical equation reflect the fact that atoms are
neither created nor destroyed in a chemical
reaction.
2H2(g) +O2(g)--- 2H2 O
These coefficients can be interpreted in terms of
mole. In the above equation we can say that 1
mole of oxygen reacts with two moles of
hydrogen to form 2 moles of water.
Mole to mole calculations
• This procedure is based on two principles:
• 1. the number of atoms of each element is
conserved in a chemical reaction.
• 2. the coefficients in a given chemical
equation give the relative numbers of
moles of each reactant and product.
These numbers of moles are used to
construct conversion factors.
Class Practice
• Find the number of moles of N2 needed to
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produce 5.0 mol NH3 by reaction with H2.
Step 1. Write the balanced chemical equation:
N2(g) +3H2(g)---- 2NH3(g)
Step 2. It follows from the equation that 1 mol
of N2 approx. equals to 2mol of NH3 so the mol
ratio is
Substance required/ substance given =1mol
N2/2mol NH3.
• Step 3. We now apply this conversion
factor to the information given and obtain
the information required:
• Moles of N2 =(5.0 mol NH3)x1mol N2/2mol
NH3 =2.5 mol of nitrogen.
• How many moles of NH3 molecules can be
produced from 2.0 mol H2 in the same
reaction as in the previous reaction if all
the hydrogen reacts.
• ANS 1.3mol ammonia
Mass to mass calculations
• By using molar masses we can also find
the mass of one substance that a given
mass of another substance can produce.
We first convert the given mass to the
corresponding number of moles by using
the molar mass of that substance. We
then convert the number of moles of the
required substance to mass by using its
molar mass.
STEPS
• 1. Convert the given mass of one substance (in
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grams) to number of moles by using its molar
mass.
Number of moles of substance A=mass of A in
grams/molar mass of A (grams per mole)
2. Write the balanced equation for the reaction.
3. Use the mol ratio to convert from the given
number of moles of one substance to the
number of moles of the other substance.
Substance required/substance given
• 4. Convert from number of moles of the
second substance to mass (in grams) by
using the molar mass of the substance.
• Mass of B required (grams) =number of
moles of B x molar mass of B (grams per
mole)
• mB = nB x MB
Class Practice
• Calculate the mass of oxygen needed to
react with 0.450 g of hydrogen gas in the
reaction 2H2(g) + O2 (g)-- 2H2O(l).
Reaction yield
• Example:
• A major source of atmospheric carbon
dioxide is the burning of gasoline in
automobiles, so we need to know how
much CO2 is likely to be produced from a
liter of gasoline.
• 2C8 H18(l) + 25O2(g)---16CO2(g) + 18H2O
• Theoretical yield is the maximum mass of
product that can be obtained from a given
mass of reactant. The theoretical yield of
carbon dioxide is based on the
assumptions that this is the only reaction
taking place and that every molecule of
octane is converted into carbon dioxide
and water.
Class Practice
• Calculate the theoretical yield of carbon
dioxide when 702 g of octane(C8H18) is
burned.
• Steps:
• 1. Convert mass to moles of C8H18
molecules then moles of C8H18 to moles of
carbon dioxide, then moles of carbon
dioxide to mass of carbon dioxide.
Home work
• Page 166
• # 4.2,4.3,4.8,4.10,
Percentage yield
• The percentage yield is the percentage of
the theoretical yield actually achieved.
• When 24 g of potassium nitrate was
heated with lead, 13.8 g of potassium
nitrite was formed in the reaction
Pb(s) + KNO3 (s)--- PbO(s) +KNO2(s).
Calculate the percentage yield of
potassium nitrite.
Ans 68.3%
Limiting reactant
• Limiting Reactant - The reactant in a chemical
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reaction that limits the amount of product that
can be formed. The reaction will stop when all
of the limiting reactant is consumed.
Excess Reactant - The reactant in a chemical
reaction that remains when a reaction stops
when the limiting reactant is completely
consumed. The excess reactant remains
because there is nothing with which it can react.
• 8car bodies + 48 tires -- 8cars with tires + 16
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tires in excess.
No matter how many tires there are, if there are
only 8 car bodies, then only 8 cars can be
made.
Likewise with chemistry, if there is only a certain
amount of one reactant available for a reaction,
the reaction must stop when that reactant is
consumed whether or not the other reactant has
been used up.
• Example Limiting Reactant Calculation:
• A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the
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limiting reactant and how much excess reactant remains after the reaction
has stopped?
First, we need to create a balanced equation for the reaction:
4 NH 3(g) + 5 O2(g) ---4 NO (g) + 6 H 2 O (g)
Next we can use stoichiometry to calculate how much product is produced
by each reactant. .
• 2.0 g of NH3 x 1mol of NH3 / 17.0 g of NH3 x 4mol NO/4mol NH3 x 30.0g of
NO/1 mol NO =3.53 g
• 4.0 g of O2x 1 mol O2 /32.0 g of O2 x 4 mol NO/5 mol O2 x 30.0 g NO/1mol
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NO=3.00 g of NO
• The reactant that produces the lesser
amount of product in this case the
oxygen.
• Next, to find the amount of excess
reactant, we must calculate how much of
the non-limiting reactant (oxygen) actually
did react with the limiting reactant
(ammonia).
• 4.0g of O2 x 1 mol of O2 /32 g of O2 x 1mol NH3
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/5 mol O2 x 17 g NH3/1mol NH3=1.70 g of NH3
1.70 g is the amount of ammonia that reacted,
not what is left over.
To find the amount of excess reactant
remaining, subtract the amount that reacted
from the amount in the original sample.
2.00g of NH3 – 1.70 g (reacted) =0.30 g
remaining.
Class Practice
• In the synthesis of ammonia which is the
limiting reactant when 100 kg of hydrogen
reacts with 800 kg of nitrogen?
Empirical formula
• Empirical Formula - A formula that gives the simplest
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whole-number ratio of atoms in a compound.
Steps for Determining an Empirical Formula
1. Start with the number of grams of each element,
given in the problem.
2. Convert the mass of each element to moles using the
molar mass from the periodic table.
Divide each mole value by the smallest number of moles
calculated.
Round to the nearest whole number. This is the mole
ratio of the elements and is represented by subscripts in
the empirical formula.
• If the number is too far to round (x.1 ~ x.9), then multiply each
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solution by the same
factor to get the lowest whole number multiple.
e.g. If one solution is 1.5, then multiply each solution in the
problem by 2 to get 3.
e.g. If one solution is 1.25, then multiply each solution in the
problem by 4 to get 5.
Once the empirical formula is found, the molecular formula for a
compound can be determined if the molar mass of the compound is
known.
Simply calculate the mass of the empirical formula and divide the
molar mass of the compound by the mass of the empirical formula
to find the ratio between the molecular formula and the empirical
formula. Multiply all the atoms (subscripts) by this ratio to find the
molecular formula.
• A compound was analyzed and found to contain
13.5 g Ca, 10.8 g O, and 0.675 g H. What is the
empirical formula of the compound?
• Start with the number of grams of each element, given
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in the problem.
Convert the mass of each element to moles using the
molar mass from the periodic table.
Divide each mole value by the smallest number of moles
calculated. Round to the nearest whole number.
In this we divide by the moles of Ca obtained.
• This is the mole ratio of the elements and
is represented by subscripts in the
empirical formula.
• CaO2H2 = Ca(OH)2
Homework
• Page 168
• # 4.34, 4.36
Solutions
• Molarity : Molarity is the number of moles
of solute dissolved in one liter of solution.
The units, therefore are moles per liter,
specifically it's moles of solute per liter
of solution.
• molarity = amount of solute (moles)
volume of solution (liter)
Class Practice
• Calculate the molarity of glucose in a
solution made by dissolving 1.368 g of
glucose (C6 H12 O6) in enough water to
make 50.00 ml of solution.
Dilution
• The most important thing to remember concerning dilution is that
you are only adding solvent. You are not adding solute when you
dilute.
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Therefore:
moles of solute before dilution = moles of solute after dilution
or Number of millimoles of solute before = millimoles of solute after
or
Number of grams of solute before dilution = number of grams of
solute after dilution
• Since the definition for Molarity is:
• Molarity = moles of solute / volume of solution in liters
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Solving for moles of solute gives:
moles of solute = M x V of solution in liters
or
millimoles of solute = M x V of solution in ml
If Moles of solute before dilution = moles of solute after dilution
then M x V in liters before dilution = M x V in liters after dilution
or
M1V1 = M2V2
where M1 = Molarity before dilution
V1 is volume of solution before dilution
M2 = Molarity of solution after dilution
V2 = Volume of solution after dilution
We can use Molarity or mass % as the concentration term so you could have the
following alternative where mass % is used:
mass % x grams solution before dilution = mass % x grams of solution after dilution
Homework
• Page 169
• # 4.40,