Stoichiometry
Mar 26, 2015
Stoichiometry
• Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions.
• It could be amounts of substances before the reaction or amount of material that is produced by the reaction.
• Stoichiometry is all about amounts.
What a balanced chemical equation tells us
• A chemical equation represents the mole ratio of reactants and products
• CuCl2 + 2AgNO3 2AgCl + Cu(NO3)2
1mol 2mol 2mol 1mol • NaOH + HCl NaCl + H2O 1mol 1mol 1mol 1mol• 2CO + O2 -> 2CO2
2mol 1mol 2mol• Mg + 2HCl -> MgCl2 + H2
1 mol 2 mol 1 mol 1 mol
• Stoichiometry will tell you that if you have ten million atoms of sodium (Na) and only one atom of chlorine (Cl) you can only make one molecule of sodium chloride (NaCl). Nothing you can do will change that. Like this:
• 10,000,000 Na + 1 Cl --> NaCl + 9,999,999 Na
Limiting reagent• A reagent is another
word for reactant• A reaction will stop
when one reactant is used up before the other
• This is called a limiting reagent
• The other reactant is the excess reagent
• Blue is the limiting reagent
• Red is the excess reagent
Limiting reagent
2H2 + O2 2H2O
Initial mol 8 6 0
Reaction mol 8 4 8
Final mol 0 2 8
Mass-Mass stoichemtry
• Involves solving a problem in which the mass of a reactant or a product is given
Calculating the mass of a substance given the mass of another reactant
or product1. Write a balanced chemical equation2. Identify known and unknown quantities of
substances3. Calculate the number of moles of known
quantities4. Find the molar ratio and use this to calculate the
number of moles of a required substanceMol of required substance = Molar ratioMol of given substance5. Calculate the quantity o the required, unknown,
substance
Calculations• Calculate the number of moles of CO2 formed
in the combustion of ethane C2H6 in a process when 35.0 mol of O2 is consumed.The reaction is:
• 2 C2H6 + 7 O2 4 CO2 + 6 H2O
• 4 CO2 : 7 O2
• 35.0 mol O2 x 4 CO2 mol = 20.0 mol CO2
• 7 mol O2 mol
• Two moles of Mg and five moles of O2 are placed in a reaction vessel, and then the Mg is ignited according to the reaction Mg + O2 MgO.
Identify the limiting reagent in this experiment.
• 2 Mg + O2 2 MgO
2 mol 1mol (I have 5, 5 -1 = 4 mol unused)• Four moles of oxygen will remain unreacted. • Oxygen is the excess reagent, and Mg is the limiting
reagent.
• What mass of iron (III) oxide is formed from the complete combustion of 183.5g of pyites (FeS2)
• FeS2 + O2 Fe2O3 + SO2
1. 4FeS2 + 11O2 2Fe2O3 + 8SO2
2. molar mass of FeS2 = 55.85 + (2 x 32.1)
= 120.05g/mol
= 183.5g x 1 mol
120.05g
= 1.529 mol
3. Find the molar ratio of 4FeS2 to 2Fe2O3
4:2 2:1
Mol of Fe2O3 = 2 FeS2 = 1
4 2
= 1.529 mol
2
Mol of Fe2O3 = 0.7643 mol4. Molar mass of Fe2O3 = 2 x 55.85 + 3 x 16
= 159.7g/mol
0.7643 mol x 159.7g
1mol
= 122.1g of Fe2O3
• Calculate the mass of water produced when 2.8g of CH4 is burnt in the air
1. CH4 + 2O2 2H2O + CO2
2. Molar mass = 12 + 4 = 16g/mol
Mol of CH4 = 2.8g x 1 mol
16g
Mol of CH4 = 0.175 mol
3. Molar ratio
CH4 : H2O 1:2
CH4 = 1 = 0.175 mol = 0.0875 mol of H2O
H2O 2 2
4. Molar mass of H20 = 2+16 = 18g/mol
Mass of H20 = 0.0875 mol x 18g = 1.575g
1 mol
Limiting reagent • How much precipitate can you make
with only 2.6mol of KCl?
• AgNO3 + KCl • AgNO3(aq) + KCl(aq) AgCl (s) + KNO3(aq)
2.6mol
• KCL is the limiting reagent we only have a certain amount of it
1. Balance equation
AgNO3(aq) + KCl(aq) AgCl (s) + KNO3(aq)
1. Determine molar ratio
2. KCl to AgCl is 1:1
I have 2.6mol of KCl I have 2.6 mol of AgCl
Molar mass of AgCl = 143.32g/mol
2.6mol KCl x 1mol of AgCl x 143.32g of AgCl
1 mol of KCL 1mol of AgCl
= 372.63g of AgCl
If 6.3g of S8 react with Pb, how many grams of PbS are formed?
• Pb + S8 PbS• 8Pb + S8 8PbS 6.3g ?Molar mass of S8 = 32.06 x 8 = 256.48g/mol Mol of S8 = 6.6g x 1mol 256.48g = 0.025 molMolar ratio = PbS 8 S8 1Mol of PbS = 8 x 0.025 = 1.647 molMolar mass of PbS = 239.27g/molMass of PbS = 1.647mol x 239.27g 1 mol = 394.06 g
Determine the limiting reagent
• What mass of liquid water is formed when 2.3g of H2 gas and 4.55g of O2 gas react together
1.2H2 + O2 2H2O
2.2.3g 4.55g
Molar mass of H2 = 2g/mol
Mol of H2 = 2.3g x 1mol = 1.15mol
2g
Molar mass of 02 = 32g/mol
Mol of O2 = 4.55 x 1 mol = 0.142mol
32g
2H2 + O2 2H2O
1.15 0.142
0.142 x 2 = 0.284mol of H2
Which is the limiting reagent?02
2H2 + O2 2H2O
0.284 0.142 0.284
Molar mass of H2O = 18g/mol
Mass of H2O = 0.282mol x 18g
1 mol
= 5.076g
What volume of 0.100mol/L H2SO4 acid reacts completely with 17.8mL
of 0.15 of potassium hydroxide?
Acid + base salt + waterKOH + H2SO4 K2SO4 + H202KOH + H2SO4 K2SO4 + 2H20
0.15M 0.1M17.8mL ?(0.0178L)Mol of KOH = 0.0178L x 0.15mol 1L = 0.00267mol•
• Molar ratio = H2SO4 1• KOH 2• Mol of H2SO4 = ½ x
0.00267mol• L of H2SO4 =
0.00136mol x 1L
0.1mol
• L of H2SO4 = 0.01335L
Zinc metal is reacted with 400mL of a 0.25mol/L solution of sulfuric acid. Calculate the mass of zinc
sulfate formed
Questions
• Pg 263 Q 1, 2, 4, 6,7,8
• Pg 267 Q 9
Titration• A titration is a method of analysis that will
allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction.
• http://www.wesleylearning.ie/resources/science/chemistry/experiments/ethanoic_acid_vinegar/index.htm