Finally, it’s Wednesday

• ARIS homework available tomorrow • Quiz Tuesday over end of chapter 3 • Exam II in two weeks

A process in which one or more substances is changed into one or more new substances is a

chemical reaction

A

chemical equation

uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H 2 with O 2 to form H 2 O reactants products 3.7

How to “Read” Chemical Equations

2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO

IS NOT

2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7

Balancing Chemical Equations

1. Write the

correct

formula(s) for the reactants on the left side and the

correct

formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2. Change the numbers in front of the formulas (

coefficients

) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6

NOT

C 4 H 12 3.7

Balancing Chemical Equations

3. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left C 2 H 6 + O 2 1 carbon on right 2 CO 2 + H 2 O multiply CO 2 by 2 6 hydrogen on left C 2 H 6 + O 2 2 hydrogen on right 2 CO 2 + 3 H 2 O multiply H 2 O by 3 3.7

Balancing Chemical Equations

4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2 CO 2 + 3 H 2 O multiply O 2 by 7 2 2 oxygen on left C 2 H 6 + O 2 2 C 2 H 6 + 7 O 2 4 oxygen + 3 oxygen = 7 oxygen (2x2) (3x1) on right 2 CO 2 + 3 H 2 O remove fraction multiply both sides by 2 4 CO 2 + 6 H 2 O 3.7

Balancing Chemical Equations

5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2 C 2 H 6 + 7 O 2 4 C ( 2 x 2) 12 H ( 2 x 6) 14 O ( 7 x 2) 4 CO 2 + 6 H 2 O 4 C 12 H ( 6 x 2) 14 O ( 4 x 2 + 6 ) Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O 3.7

Amounts of Reactants and Products

1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units 3.8

Channel Setting Instructions for

ResponseCard RF

ALEX – SELECT PARTICIPANT LIST

1. Press and release the "GO" button.

2. While the light is flashing red and green, enter the 2 digit channel code (i.e. channel 1 = 01, channel 21 = 21).

3. After the second digit is entered, Press and release the "GO" button.

4. Press and release the "1/A" button. The light should flash yellow to confirm.

How many moles of carbon dioxide are formed during the combustion of 3 moles of ethane?

1. 3 moles

25% 25% 25% 25%

2. 4 moles 3. 5 moles 4. 6 moles

3 m ol es 4 m ol es 5 m ol es 6 m ol es

Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH 3 OH moles CH 3 OH molar mass CH 3 OH moles H coefficients chemical equation 2 O grams H molar mass H 2 O 2 O 209 g CH 3 OH x 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O = 235 g H 2 O 3.8

Consider the combustion of glucose: C 6 H 12 O 6 + O 2 → CO 2 + H 2 O How many grams of water will be formed from 10 grams glucose?

1. 60 g

25% 25% 25% 25%

2. 6 g 3. 3 g 4. 12 g MW C 6 H 12 O 6 – 180.18 g/mol

6 0 g 6 g 3 g 1 2 g

Limiting Reagents 2NO + 2O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent 3.9

Balance the following reaction NH

3

+ CO

2

→ (NH

2

)

2

CO + H

2

O

1. 2, 1, 1, 1 2. 4, 2, 2, 6 3. 3, 1, 1, 2 4. 5, 3, 2, 1

25% 25% 25% 25% 2 , 1 , 1 , 1 4 , 2 , 2 , 6 3 , 1 , 1 , 2 5 , 3 , 2 , 1

Do You Understand Limiting Reagents?

g Al g Fe 2 In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed.

O 3 mol Al mol Fe 2 O 3 mol Fe 2 O 3 needed OR mol Al needed g Fe 2 O 3 needed g Al needed 124 g Al x 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 = 367 g Fe 2 O 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 3.9

Use limiting reagent (Al) to calculate amount of product that can be formed.

g Al mol Al 2Al + Fe 2 O 3 mol Al 2 O 3 Al 2 O 3 + 2Fe g Al 2 O 3 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 = 234 g Al 2 O 3 3.9

Reaction Yield

Theoretical Yield

is the amount of product that would result if all the limiting reagent reacted.

Actual Yield

is the amount of product actually obtained from a reaction.

% Yield

= Actual Yield Theoretical Yield x 100 3.10

The reaction yield may not go above 100%. This is in accordance with the

1. Law of definite proportions

25% 25% 25% 25%

2. Law of multiple proportions 3. Law of conservation of mass 4. Unwritten book of the road

L aw o f d ef in ite L p aw ro o po f m ...

ul tip le L p ro aw po o ...

f c on se rv U at nw io ri n tt ..

en b oo k of th ...

Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3H 2 (

g

) + N 2 (

g

) 2NH 3 (

g

) NH 3 (

aq

) + HNO 3 (

aq

) NH 4 NO 3 (

aq

) fluorapatite 2Ca 5 (PO 4 ) 3 F (

s

) + 7H 2 SO 4 (

aq

) 3Ca(H 2 PO 4 ) 2 (

aq

) + 7CaSO 4 (

aq

) + 2HF (

g

)