I am a nice shark, not a mindless eating machine. If I am to change this image, I must first change myself. Fish are friends, not food.I am a nice shark, not a mindless eating machine. If I am to change this image, I must first change myself. Fish are friends, not food.” - from “Finding Nemo”

IN OUR PREVIOUS CLASS ABOUT CHEMICAL EQUATIONS, WE FOUND THAT AN EQUATION CAN BE USED TO DESCRIBE A CHEMICAL REACTION. IT GIVES US A GREAT DEAL OF INFORMATION: CH 4 + 2O 2  CO 2 + 2H 2 O 1 MOLECULE 2 MOLECULES METHANE OXYGEN CARBON DIOXIDE WATER 1 MOLE 2 MOLES 16 g 64 g 44 g 36 g 6.02 x (6.02x10 23 ) 6.02 x (6.02 x ) Molecules molecules molecules molecules

THE MOST USEFUL RELATIONSHIIPS WILL BE THE MOLAR RATIOS INVOLVED IN THE REACTIONS AND THE MASS RATIOS. WE CAN USE THIS INFORMATION TO CALCULATE: 1.THE AMOUNTS OF STARTING MATERIALS NEEDED TO MAKE A GIVEN AMOUNT OF PRODUCT 2.THE THEORETICAL YIELD (100%) THAT GIVEN AMOUNTS OF PRODUCTS WOULD GIVE 3.THE LIMITING REAGENT IN A REACTION

HOW MANY GRAMS OF WATER ARE PRODUCED IN THE OXIDATION OF 1 GRAM OF GLUCOSE? C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O (combustion reaction) 1 MOLE 6 MOLES GLUCOSE WATER

LET’S CALCULATE THE MOLAR MASSES OF WATER AND GLUCOSE SO WE CAN GET THE MASS RELATIONSHIPS. The molar mass of glucose is: 6 C = 6 x 12.0 = 72.0 g 12 H = 12 x 1.01 = g 6 O = 6 x 16.0 = 96.0 g molar mass = g/mole The molar mass of water is: 2 H = 2 x 1.01 = 2.02 g 1 ) = 1 x 16.0 = 16.0 g molar mass = g/mole

1.0 g glucose X g water C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O 180 g/mole 6(18 g/mole) 1/180 = X/108 OR X = 108/180 = 0.6 grams WATER

The decomposition of KClO 3 (potassium chlorate) is commonly used to prepare small amounts of O 2 in the laboratory: 2 KClO 3 (s)  2 KCl (s) + 3 O 2 (g) How many grams of O 2 can be prepared from 4.50 g KClO 3 ?

FIRST, THE FORMULA MASS OF KCLO 3 IS: 1 K = 1 x 39.1 = 39.1 g 1 Cl = 1 x 35.5 = 35.5 g 3 O = 3 x 16.0 = 48.0 g formula mass = 123 g/mole THE FORMULA MASS OF O 2 IS 2 X 16 = 32 g/mole 4.5 gX g 2 KClO 3  2 KCl + 3 O 2 2(123 g/mole)3(32 g/mole)

SO, YOUR RATION IS: 4.5/246 = X/96 AND X = (4.5/246) x 96 = 1.76 g OXYGEN 2 KClO 3 (s)  2 KCl (s) + 3 O 2 (g)

SOLID LITHIUM HYDROXIDE IS USED IN SPACE VEHICLES TO REMOVE EXHALED CARBON DIOXIDE. THE LITHIUM HYDROXIDE REACTS WITH GASEOUS CARBON DIOXIDE TO FORM SOLID LITHIUM CARBONATE AND LIQUID WATER. HOW MANY GRAMS OF CARBON DIOXIDE CAN BE ABSORBED BY 1.00 G OF LITHIUM HYDROXIDE? YOU NEED THE BALANCED EQUATION, SO 2 LiOH (s) + CO 2  Li 2 CO 3 (s) + H 2 O (l)

The formula mass of LiOH is: 1 Li = 1 x 6.94 = 6.94 g 1 O = 1 x 16.0 = 16.0 g 1 H = 1 x 1.01 = 1.01 g formula mass = 24.0 g and you have 2, so 2 x 24.0 = 48.0 g The molar mass of CO 2 = x 16.0 = 44.0 g 1.0 g X g 2 LiOH (s) + CO 2  Li 2 CO 3 (s) + H 2 O (l) 48 g 44.0 g So, 44 / 48 = X / 1.00 and X = (44/48) x 100 X = g