Stoichiometry

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Stoichiometry - PowerPoint PPT Presentation

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.

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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 product
• Write a balanced chemical equation
• Identify known and unknown quantities of substances
• Calculate the number of moles of known quantities
• Find the molar ratio and use this to calculate the number of moles of a required substance

Mol of required substance = Molar ratio

Mol of given substance

5. 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
• 4FeS2 + 11O2 2Fe2O3 + 8SO2
• 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 mol

4. 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
• CH4 + 2O2 2H2O + CO2
• 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
Balance equation

AgNO3(aq) + KCl(aq) AgCl (s) + KNO3(aq)

• Determine molar ratio
• 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

• 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 mol

Molar ratio = PbS 8

S8 1

Mol of PbS = 8 x 0.025 = 1.647 mol

Molar mass of PbS = 239.27g/mol

Mass 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
• 2H2 + O2 2H2O
• 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 + water

KOH + H2SO4 K2SO4 + H20

2KOH + H2SO4 K2SO4 + 2H20

0.15M 0.1M

17.8mL ?

(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.htmhttp://www.wesleylearning.ie/resources/science/chemistry/experiments/ethanoic_acid_vinegar/index.htm