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StoichiometryStoichiometryMoles, masses, reactions, oh my!
StoikheionStoikheion + + metriametriaFrom the Greek for “element”
and “the process of measuring”Quantitative analysis of reactants
and productsMolesMolar massesBalanced equations
Grinch SandwichesGrinch SandwichesIngredients per sandwich
◦Three pieces of bread◦Sauerkraut (10 g)◦Toadstool (15 g)◦Arsenic sauce (5 g)
Balanced EquationsBalanced Equations2H2 + O2 2H20
◦Coefficients = moles◦2 moles of hydrogen react with 1 mole
of oxygen to produce 2 moles of water◦What if you only had 1 mole of
hydrogen?◦What if you had 10 moles of both
hydrogen and oxygen?◦What if you needed to make 5 moles
of water?
Law of Conservation of Law of Conservation of MassMassConvert each of the moles in the
previous equation to gramsAdd the reactants’ masses
together.Compare to the product’s mass.Be in awe!
Mole RatioMole Ratio ssEquation must be balancedRelate reactants to each other,
products to each other, or reactants to products
Molar MassesMolar MassesIf mass is mentioned, find that
substance’s molar mass
Making Stoichiometry Making Stoichiometry WorkWorkDimensional Analysis
◦Number, unit, substance…always!
Problem #1Problem #1When 2.50 g of hydrogen gas are
mixed with an excess of oxygen gas, how many grams of water should be produced?
Problem #1—step 1Problem #1—step 1Write a balanced equation.
◦2H2 + O2 2H20
Problem #1—step 2Problem #1—step 2Begin with your given information
and make a plan of conversions.◦Grams of hydrogen moles of
hydrogen moles of water grams of water
Write it as dimensional analysis.
g H2 x mol H2 x mol H2O x g H2O =
g H2 mol H2 mol H20
Problem #1—step 3Problem #1—step 3Fill in your numbers.
2.5 g H2 x 1 mol H2 x 2 mol H2O x 2.02 g H2 2 mol H2
18.02 g H2O =1 mol H20 • Cross out what divides out.• Calculate your answer.
Problem #1—step 4Problem #1—step 4
Problem #2Problem #2The thermite reaction: aluminum
foil reacts with ferric oxide
Problem #2Problem #2Wow! Now let’s assume that
0.010 g of aluminum reacted with the excess ferric oxide on the ball bearing. What mass of aluminum oxide should be produced?
Problem #2—Step 1Problem #2—Step 1Write a balanced equation.
2Al + Fe2O3 Al2O3 + 2Fe
Problem #2—Step 2Problem #2—Step 2Write your plan.
g Al x mol Al x mol Al2O3 x g Al2O3 =
g Al mol Al mol Al2O3
Problem #2—Step 3Problem #2—Step 3Place the proper numbers and
calculate.
0.010 g Al x 1 mol Al x 1 mol Al2O3 x
26.98 g Al 2 mol Al101.96 g Al2O3 =
1 mol Al2O3
Problem #2—Answer Problem #2—Answer 0.0189 g of Al2O3 should be
formed
Problem #3Problem #3What if only 0.0150 g of Al2O3
were actually produced in the previous problem? What is the percent yield?
So, you can calculate the % yield◦% yield = actual amt x 100
theoretical amt
Problem #3Problem #3Thus, the % yield calculation
0.0150 g x 100 = 79.4 % yield0.0189 g
Limiting ReagentLimiting ReagentIdeally, the perfect amount of
each reactantOften one reactant limits how
much product can be formed—limiting reagent (LR)
Often the other reactant(s) is(are) in excess (ER)
You will be given at least 2 quantities of reactants
Grinch SandwichesGrinch SandwichesIngredients per sandwich
◦Three pieces of bread◦Sauerkraut (10 g)◦Toadstool (15 g)◦Arsenic sauce (5 g)
Problem #4Problem #40.500 g of silver nitrate in
solution reacts with 0.750 g of tin (IV) chloride in solution. If 0.400 g of solid silver chloride are retrieved, what is the percent yield?
Problem #4—Step 1Problem #4—Step 1Write a balanced equation.
4AgNO3 + SnCl4 4AgCl + Sn(NO3)4
Write shorthand, to save time. You’ll write the real stuff in the answer.
A B C D
Problem #4—Step 2Problem #4—Step 2Ask yourself, when I have 0.500 g
of AgNO3, how many grams of SnCl4 do I really need? This will determine the LR.
Set up the plan
g A x mol A x mol B x g B = g A mol A mol B
Problem #4—Step 3Problem #4—Step 3Plug in the numbers.
0.5 g A x 1mol A x 1 mol B 169.88g A 4 mol A
x 260.50 g B = 0.192 g SnCl4 1 mol A
Problem #4—Step 4Problem #4—Step 4Analyze and determine the LR
◦0.750 g of SnCl4 are available, and only 0.192 g of it need to be used to completely react with the AgNO3
•Which substance will be in excess?•Which substance will you run out
of first?
Problem #4—Step 5Problem #4—Step 5
• Now, use the given amount of the LR to find how much product you should be able to make.• Set up your plan.
g A x mol A x mol C x g C = g A mol A mol C
Problem #4—Step 6Problem #4—Step 6
• Plug in your numbers and calculate.
0.5 g A x 1 mol A x 4 mol C x 169.88 g A 4 mol A
143.32 g C = 0.422 g AgCl should be
1 mol C formed
Problem #4—Step 7Problem #4—Step 7
• Look at the actual amount of product that was formed, and determine percent yield using the theoretical you just calculated.
0.400 g AgCl x 100 = 94.8% yield0.422 g AgCl
PracticePracticeSit down with some bread,
sauerkraut, toadstools, and arsenic sauce, and work some stoichiometry problems just for fun.