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### 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

- Copper + sulfur Copper(II) sulfide
- Calcium oxide + water Calcium hydroxide
- Hydrogen + nitrogen
- Chromium + oxygen
- Aluminum + bromide
- Sodium + iodine
- 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

1 u = weight of 1/12 of 12C

2Na + I2 2NaI

- 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 1023atoms of Na and 1 mole of Na

254 g of I2 (formula weight = 254u) correspond to 6.02 x 1023molecules of I2 and 1 mole of I2

150 g of NaI (formula weight = 150u) correspond to 6.02 x 1023molecules of NaI and 1 mole of NaI

Avogadro’s Number

Number of atoms

How many atoms are in 10 g of sodium?

How many molecules in 20 g of iodine?

= 26.3 x 1022 atoms

6.02 x 1023 atoms

6.02 x 1023 molecules

10 g x

20 g x

= 2.63 x 1023 atoms

= 4.74 x 1022 molecules

23 g

254 g

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 gof 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

- Find the formula weights of the compounds involved in the reaction
- Calculate the number of moles given by the actual mass of reagents
- Calculate the number of moles of reagents used in the reaction
- Calculate the number of moles of product produced
- 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

Next class

- Solutions and numerical representation.

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