Whats it mean? Greek stoikheion, meaning elementelement metron,
meaning measuremeasure In English.chemical recipe
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IDEAL STOICHIOMETRY Assumes a reaction goes to completion and
that all reactants will turn into products.
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STEPS to SOLVING PROBLEMS 1. Write and balance the chemical
reaction. The coefficients will give the mole ratio (the recipe).
2. Find the question or unknown. ?____ = given x conversion x mole
mole 3. Start with the given and use conversions to reach the
unknown. Every problem will include a mole ratio step.
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Go with the flow. chart to solve the 4 simple types of problems
^ (use coefficients from balanced reaction) Grams of given Molar
mass Moles given Mole ratio Moles unknown Molar mass Grams
unknown
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1. MOLE MOLE What are we calculating? Moles of any product or
reactant, given moles of one other product or reactant.
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EX. How many moles of ammonia are produced when 6 moles of
hydrogen gas react with an excess of nitrogen gas? Balanced
reaction: 3H 2 (g) + N 2(g) 2NH 3 (g) ^moles ^mole ^moles ? Moles
NH 3 = 6 moles H 2 x 2 moles NH 3 = 4 moles NH 3 1 3 moles H 2
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Ex. How many moles of nitrogen are needed for the reaction to
go to completion? ? Moles N 2 = 6 moles H 2 x 1 mole N 2 = 2 moles
N 2 1 3 moles H 2
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2. MOLE - MASS What are we calculating? Grams of any product or
reactant, given moles of any product or reactant.
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Ex. When magnesium burns in air, it combines with oxygen to
form magnesium oxide. What mass of magnesium oxide is produced from
2.00 moles of magnesium? Balanced Reaction: 2 Mg (s) + O 2 (g) 2MgO
(s) ?g MgO = 2 moles Mg x 2 moles MgO x 40.31 g MgO = 80.6 g MgO 1
2 moles Mg 1 mole MgO
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3. MASS MOLE What are we calculating? Moles of any product or
reactant, given grams of any product or reactant. Ex. How many
moles of mercury (II) oxide are needed to produce 125 grams of
oxygen in a decomposition reaction? Balanced Reaction: 2 HgO (s) O
2 g) + 2Hg (l) ?moles HgO= 125g O 2 x 1mole O 2 x 2 moles HgO =
7.81 moles HgO 1 32.00 g O 2 1 mole O 2
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4. MASS - MASS What are we calculating? Grams of any product or
reactant, given grams of any product or reactant. Ex. Nitrous oxide
is sometimes used as an anesthetic in dental work. It is produced
when ammonium nitrate is decomposed into dinitrogen monoxide and
water. How many grams of ammonium nitrate are needed to produce
33.0 g of N 2 O? Balanced Reaction: NH 4 NO 3(s) N 2 O (g) + 2H 2 O
(l) ?g NH 4 NO 3 = 33.0g N 2 O x 1 mole N 2 O x 1 mole NH 4 NO 3 x
80.06g NH 4 NO 3 = 60.0g NH 4 NO 3 1 44.02 g N 2 O 1 mole N 2 O 1
mole NH 4 NO 3
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How many grams of water are produced during the reaction? (Law
of Conservation = 60.0 g 33.0 g) ?g H 2 O = 33.0 g N 2 O x 1 mole N
2 O x 2 moles H 2 O x 18.02 g H 2 O = 27.0 g H 2 O 1 44.02 g N 2 O
1 mole N 2 O 1 mole H 2 O
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MORE EXAMPLES. Acetylene gas (C2H2) is used in welding and
produces a very hot flame when burned in pure oxygen. How many
grams of each product are produced when 25 kg of acetylene burns
completely? Balanced Reaction: C 2 H 2 + 5/2 O 2 2 CO 2 + H 2 O 2C
2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O ? gCO 2 = 25kg x 10 3 x 1 moleCO 2 x
4 moleCO 2 x 44.01gCO 2 = 85000 g CO 2 1 1kg 26.04g C 2 H 2 2 moleC
2 H 2 1 mole CO 2 ?g H 2 O= 25kg x 10 3 x 1 moleCO 2 x 2 mol H 2 O
x 18.02 g H 2 O = 1.7 x 10^4g 1 1kg 26.04g C 2 H 2 2 moleC 2 H 2 1
mole H 2 O H 2 O
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MOLAR VOLUME of a GAS Rather than weighing gaseous products and
reactants, it is often easier to consider the VOLUME of gas reacted
or produced. For this to be useful in a stoichiometry problem, we
must relate the volume of a gas to a number of moles. MOLAR VOLUME
OF A GAS Every gas has volume = 22.4 L at STP STP = standard
temperature and pressure 1 atm and 0.0 degrees C
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Ex. Using molar volume Acetylene burns in oxygen. If 100.0 L of
acetylene is burned at STP, what volume of carbon dioxide is
produced? 2 C 2 H 2 + 5 O 2 4 CO 2 + 2H 2 O ?L CO 2 = 100.0L C 2 H
2 x1 mol C 2 H 2 x 4 mol CO 2 x22.4 L CO 2 = 200.0 1 22.4LC 2 H 2 2
mol C 2 H 2 1 mol CO 2 L CO 2 How many moles of water will be
produced? ?mol H 2 O = 100.0L C 2 H 2 x1 mol C 2 H 2 x 2 mol H 2 O
= 4.464 mol H 2 O 1 22.4 LC 2 H 2 2 mol C 2 H 2
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If 84 L of acetylene are used, how many liters of oxygen are
needed for complete combustion? ? L O 2 = 84 L C 2 H 2 x 1 mol C 2
H 2 x 5 mol O 2 x 22.4L O 2 = 210 L O 2 1 22.4 L C 2 H 2 2 molC 2 H
2 1 mol O 2
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PERCENT YIELD and STOICHIOMETRY THEORETICAL YIELD The maximum
amount of product that can be produced during a reaction. Predicted
by stoichiometry and ideal conditions. ACTUAL YIELD How much
product is really produced during the reaction. A percent yield
calculation tells us how efficient a reaction is.
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% YIELD % yield = actual yield x 100 theoretical % yield + %
error = 100%
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% yield examples A reaction is supposed to produce 200. grams
of HCl. If it actually yield 178 grams of HCl, how efficient is the
reaction? % yield = actual x 100 = 178g x 100 = 89.0 %yield ; 11 %
error theoretical 200g
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If chlorine and hydrogen are combined to form HCl, if you are
to obtain a 98% yield and you start with 3.0 moles of hydrogen,
what amount of HCl is produced? Cl 2 + H 2 2HCl 3.0mol 98% = A
(220)98% = A (220) A= (220)(.98) T 220 A=215.6gHCl What is the %
yield of HCl if you start with 5.0 moles of hydrogen and obtain 345
grams of HCl? 345g x 100 = 345 x 100 = 95.8% T 360 ?g HCl = 5 mol H
2 x 2 mol HCl x 36.46 g HCl = 364.6g HCl 11 mol H 2 1 mol HCl 360 g
HCl (sigfigs)
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LIMITING REAGENT LIMITING REACTANT We have already seen that
reaction conditions are not ideal limit the amount of product
produced when dealing with % yield calculations. Now suppose one
reactant is completely consumed during the reaction? What is a
limiting reactant? The reactant that is completely consumed during
a reaction. Limits the amount of product formed. Excess reactant?
The reactant that is left over after a reaction. The amount used
depends on the limiting reactant.
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Steps for determining the LR and EXCESS reagent 1. Set up 2
stoichiometry problems beginning with the given amount of each
reactant and determine the amount of 1 product that may be formed.
2. Select the reagent that limits the amount of product by choosing
the one that produces less product. This is your limiting reagent.
3. Determine how much of the other reactant is used starting with
the limiting reagent and using stoichiometry. 4. Subtract the
amount of excess reactant used from the given amount to find the
amount left over.
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LIMITING REAGENT example Zinc and sulfur react to form zinc
sulfide. If 135.68 grams react with 66.02 grams of sulfur determine
the following: 8Zn + S 8 8ZnS a. What is the limiting reagent?
?gZnS = 135.68g Zn x 1 mol Zn x 8 mol ZnS x 97.48g ZnS = 202.20g
ZnS 1 65.41g Zn 8 mol Zn 1 mol ZnS ?gZnS = 266.02g S 8 x 1 mol S 8
x 8 mol ZnS x 97.48g ZnS = 808.59g ZnS 1 256.56g S 8 1 mol S 8 1
mol ZnS
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b. Which reactant is in excess? S 8 c. How much zinc sulfide
could be produced? 202.20g Zn d. What is the maximum amount of
product? 202.20g ZnS e. How many grams of excess reagent (S 8 )
remain? ?gS 8 = 135.68g Zn x 1 mol Zn x 1 mol S 8 x 256.56g S 8 =
66.52g S 8 used used 1 65.41g Zn 8 mol Zn 1 mol S 8 266.02g S 8
given -66.52g S 8 used 199.50g S 8 left over