Ch. 3.8 Stoichiometry
•
Stoichiometry Refresher
o Use “Mole Method”
§ Coefficients in chemical reaction indicate relative numbers of moles of each substance.
§ Ratio of one substance to the other becomes conversion factor in dimensional analysis
• Mole-Mole conversions are central to all stoichiometry problems
§ Can compare any product or reactant to any other product or reactant
§ Example:
Given the balanced chemical equation,
N2 + 3 H2 → 2 NH3
determine how many grams of hydrogen would be needed to completely react with 33.8 g N2:
• According to the equation, 1 mol N2 ≏ 3 mol H2 (≏ means “stoichiometrically equivalent”), so
get 2 conversion factors:
1 mol N 2
3 mol H 2
and
3 mol H 2
1 mol N 2
• Select the ratio that converts from moles of given to moles of unknown:
1 mol N 2 3 mol H 2 2.016 g H 2
Mass H 2 = 33.8 g N 2 ×
×
×
= 7.30 g H 2
28.02 g N 2 1 mol N 2 1 mol H 2
§ Other given units must be converted to moles first; ending units converted from moles
• Octane (C8H18) is a combustible liquid with density = 0.703 g/mL. What volume of CO2 at STP
(22.4 L/mol) can be produced from 1.00 L of octane according to the balanced equation:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
1000 mL 0.703 g 1 mol C8 H18
Volume CO2 = 1.00 L C8 H18 ×
×
×
1L
1 mL 114.2 g C8 H18
×
•
16 mol CO2 22.4 L CO2
×
= 1.10 ×103 L CO2
2 mol C8 H18 1 mol CO2
Ton-mole, kg-mol, mmol
o Can multiply both sides of gram-mol conversions to deal with very large or very small
amounts
32.00 ton O2 32.00 kg O2 32.00 mg O2
o Using O2 as an example:
;
;
; etc.
1 ton-mol O2 1 kg-mol O2 1 mmol O2
§ Use same units in mol-mol ratio between compounds
§ Don’t need to convert to/from grams unless required by problem:
• 1 ton = 1000 kg; 1 kg = 1000 g; 1 g = 1000 mg
o KClO3 is a strong oxidizing agent because when heated it decomposes into KCl and O2 gas.
What mass of O2, in mg, can be produced by decomposition of 14.8 mg of KClO3 according
to the balanced equation,
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
1 mmol KClO3
3 mmol O2 32.00 mg O2
mass O2 = 14.8 mg KClO3 ×
×
122.6 mg KClO3 2 mmol KClO3 1 mmol O2
= 5.79 mg O2
•
Homework #3-5: Problems pg. 107 #3.61, 3.66, 3.68, 3.72, 3.74, 3.76, 3.78, 3.106, 3.144