GSCI 163
Lecture 7
Review
• Elements combine in chemical reactions to
produce new substances with different
chemical properties.
• Balancing chemical equations require:
– Keeping the same number of elements on both
sides of the chemical equation by adjusting
coefficients (never subscripts of the chemical
formula)
Activity
• Determine the chemical formula (yourself) of
each substance in the following chart
• Find the appropriate chemical equation (your
partner in the reaction)
• Balance the chemical equation for each
reaction
Find your partner and write the chemical
reactions
1)
Copper + sulfur  Copper(II) sulfide
2)
Calcium oxide + water  Calcium hydroxide
3)
Hydrogen + nitrogen 
4)
Chromium + oxygen 
5)
Aluminum + bromide 
6)
Sodium + iodine 
7)
Hydrogen + chlorine 
How much is produced?
• In a chemical reaction, a certain mass of reagents
combine to produce a certain mass of products.
Consider the following reaction:
2Na + I2  2NaI
Here two atoms of sodium react with one molecule
of iodine to produce two molecules of sodium
iodide.
Problem
How much sodium iodide is produced when 10
grams of sodium react with 20 gram of iodine?
• Questions that need to be answered:
– How many atoms are there in 10 grams of Na?
– How many molecules are there in 20 g of I2?
Start with atomic weight
2Na
+ I2  2NaI
1 u = weight of
1/12 of 12C
• Atomic weight of Na = 23u, iodine = 127u.
Formula weight is the sum of the weight of the elements of
the compound
Formula weight of iodine molecule = 254u
Formula weight of sodium iodide = 150u
2 x 23 u + 254 u = 2 x 150u
300 u = 300 u
Moles and Avogadro’s number
23 g of Na (atomic weight = 23u) correspond to
6.02 x 1023 atoms of Na and 1 mole of Na
254 g of I2 (formula weight = 254u) correspond to
6.02 x 1023 molecules of I2 and 1 mole of I2
150 g of NaI (formula weight = 150u) correspond to
6.02 x 1023 molecules of NaI and 1 mole of NaI
Avogadro’s Number
Number of atoms
How many atoms are in 10 g of sodium?
10 g x
6.02 x 1023 atoms
= 2.63 x 1023 atoms = 26.3 x 1022 atoms
23 g
How many molecules in 20 g of iodine?
20 g x
6.02 x 1023 molecules
254 g
= 4.74 x 1022 molecules
Reaction
2Na + I2  2NaI
Each molecule of iodine reacts with 2 atoms of
sodium to produce 2 molecules of NaI
Thus, 4.74 x 1022 molecules will react with 9.48 x
1022 atoms of sodium, producing 9.48 x 1022
molecules of NaI
Final product
• 16.8 x 1022 atoms of Na are left out. We say
that one of the reagents (iodine) was used up.
(This corresponds to 6.4 g of Na)
To answer our initial question:
• (9.48 x 1022) x 150 g / 6.02 x 1023 = 23.6 g of
sodium iodide was produced
Can we make this simpler?
Use moles instead:
1 mole = 6.02 x 1023 atoms, molecules, etc
Thus,
• 23 g of Na is 1 mole  10 g of Na is 0.435
mole
• 254 g of I2 is 1 mole  20 g of I2 is 0.079 mole
Reaction
2Na + I2  2NaI
Each molecule of iodine reacts with 2 atoms of
sodium to produce 2 molecules of NaI
Thus, 0.079 mole of iodine will react with 0.157
(2 x 0.079) mole of sodium, producing 0.157
mole of NaI
Final product
• 0.277 mole of Na are left out (0.435 - 0.157).
We say that one of the reagents (iodine) was
used up.
(This corresponds to 6.4 g of Na)
To answer our initial question:
• 0.157 x 150 g = 23.6 g of sodium iodide was
produced
Recipe
1. Find the formula weights of the compounds involved in
the reaction
2. Calculate the number of moles given by the actual mass of
reagents
3. Calculate the number of moles of reagents used in the
reaction
4. Calculate the number of moles of product produced
5. Convert the number of moles of the product to mass
using the formula weight of the product
Exercise: calculate the mass of water produced by the
reaction of 10 g of hydrogen gas with 10 g of oxygen gas.
How we classify reactions
• Combination or synthesis X + Y  XY
• Decomposition reaction XY  X + Y
• Replacement reaction XY + Z  XZ + Y
• Ion exchange reaction
3Ca(OH)2 + Al2(SO4)3  3CaSO4 + 2Al(OH)3
• Redox reaction – Gain and losses of electrons
Types of chemical reaction
Next class
• Solutions and numerical representation.