lecture 4.4 – particles to particles and complex stoichiometry
Post on 16-Jan-2016
28 Views
Preview:
DESCRIPTION
TRANSCRIPT
Lecture 4.4 – Particles to Particles and Complex Stoichiometry
Today’s Learning Targets• LT 4.7 – I can convert from particles of one
compound to particles of another compound • LT 4.8 – I can complete a complex stoichiometric
conversion that incorporates molarity, particles, mass, moles, and volumes of substances.
Grams of Substance A
Grams of Substance B
Molarity of Substance B
Atoms of Substance
A
Moles of A
Moles of B
Atoms of Substance
B
Molarity of Substance A
Our Focus for Today
Atoms of Element or Compound A
Atoms of Element or Compound B
Moles of A
Moles of B
How do we convert particles of one substance to particles of another substance?
I. Moles to Particles Conversion Factor
• Our conversion factor for moles particles is:
6.022 x 1023 particles
mole
Can be reversed to fit conversion
II. Moles to Particles Between 2 Substances
• We can utilize what we know about converting between moles to convert between particles for a different substance.
• We simply follow the same rules and conversion factors
Class Example• You react 30 x 1023 particles of HCl for the
following reaction:HCl + MnO2 MnCl4 + H2O
• How many molecules of MnCl4 did you use for this reaction?
Example• You react 3 moles of MnO2 for the following
reaction:HCl + MnO2 MnCl4 + H2O
• How many molecules of HCl did you use for this reaction?
Table Talk• For the following reaction:
NH4SCN + FeCl3NH3(g) + HCl + Fe(SCN) 3
• If you react 4 moles of FeCl3, how many particles of NH3 would you produce?
Table Talk• If you react 12 x 1023 particles of MnO2 with
HCl in the following reaction:HCl + MnO2 MnCl4 + H2O
• How many moles of MnCl4 did you use in this reaction?
How do I complete complex stoichiometry problems?
Grams of Substance A
Grams of Substance B
Molarity of Substance B
Atoms of Substance
A
Moles of A
Moles of B
Atoms of Substance
B
Molarity of Substance A
Class Example• You wish to make mustard gas using the
reaction:(C2OH5)2S + 2 HCl → (C2H4Cl)2S + 2 H2O
• If I add 52 g of HCl, then how many particles of H2O will I produce?
Grams of Substance A
Grams of Substance B
Molarity of Substance B
Atoms of Substance
A
Moles of A
Moles of B
Atoms of Substance
B
Molarity of Substance A
Table Talk• You wish to make mustard gas using the
reaction:(C2OH5)2S + 2 HCl → (C2H4Cl)2S + 2 H2O
• If I add 143 g of (C2OH5)2S , then how many grams of (C2H4Cl)2S will I produce?
Grams of Substance A
Grams of Substance B
Molarity of Substance B
Atoms of Substance
A
Moles of A
Moles of B
Atoms of Substance
B
Molarity of Substance A
Table Talk• You run the Haber – Bosh Reaction to make
NH3:
N2 + 3H2 2NH3
• If you begin with 15 x 1023 particles of H2, then how many grams of NH3 do you produce?
Grams of Substance A
Grams of Substance B
Molarity of Substance B
Atoms of Substance
A
Moles of A
Moles of B
Atoms of Substance
B
Molarity of Substance A
top related