Al2O3  O2 + Al 2 Al2O3  3 O2 + 4 Al MOLES REVIEW, MAY 28 2 3 4 2 6 6 a) What is the mole ratio of O2 to Al? O2 = Al THE MOLE RATIOS OF SUBSTACES IN A REACTION ARE THE BALANCED REACTION COEFFICIENTS. 3 4

2 Al2O3  3 O2 + 4 Al b) If 6 mol of O2 completely reacts, how many moles of Al are produced? O2 = = Al 3 6 X = 8 MOL of Al 4 X 2 Al2O3  3 O2 + 4 Al c) If 6 mol of O2 completely reacts, how many grams of Al are produced? Moles = mass GFM 8 = mass = 215.2g Al 26.9

Al 2 Al2O3  3 O2 + 4 Al 2 O 3 101.93 g/mol X 26.98g/mol = 53.96 g d) What is the % by mass composition of Al2O3 ? NEVER USE COEFFICENTS FOR MOLAR MASS CALCULATIONS % COMP (MASS) = PART X 100 WHOLE ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS EQUALS MASS SUBTOTAL Al 2 X 26.98g/mol = 53.96 g O 3 15.99g/mol 47.97 g + 101.93 g/mol First you need to find the molar mass (gfm) of the substance

Al 2 O 3 X 26.98g/mol = 53.96 g 15.99g/mol 47.97 g 101.93 g/mol % COMP (MASS) = PART X 100 WHOLE % COMP Al = 53.96 g X 100 = 52.93 % 101.93 g ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS EQUALS MASS SUBTOTAL Al 2 X 26.98g/mol = 53.96 g O 3 15.99g/mol 47.97 g 101.93 g/mol

C 1 H 13.017 g/mol X 12.01g/mol = 12.01 g 1.007g/mol 1.007 g + REVIEW CLASS MULTIPLE CHOICE QUESTIONS, FRIDAY JUNE 1 2007 #7) CALCULATE THE MOLECULAR FORMULA OF A COMPOUND THAT IS CH, GIVEN MOLECULAR MASS IS 78 G/MOL. ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS EQUALS MASS SUBTOTAL C 1 X 12.01g/mol = 12.01 g H 1.007g/mol 1.007 g + 13.017 g/mol THIS IS THE MASS (GFM)OF THE EMPIRICAL FORMULA

REVIEW CLASS MULTIPLE CHOICE QUESTIONS, FRIDAY JUNE 1 2007 GIVEN IN PROBLEM #7)CONTINUED MOLAR MASS 78. g/mol 5.99 = 6 = = EMPIRICALMASS 13.017 g/mol THIS MEANS THAT THE MOLECULAR(TRUE) FROMULA IS 6 TIMES AS LARGE AS THE EMPIRICAL(SIMPLEST) FORMULA. MULTIPLY THE SUBSCRIPTS OF THE EMPIRICAL FORMULA BY 6. CH X 6 = C6H6

REVIEW CLASS MULTIPLE CHOICE QUESTIONS, FRIDAY JUNE 1 2007 #10) CALCULATE THE FORMULA OF A COMPOUND THAT IS 85% SILVER and 15% FLUORINE BY MASS. ELEMENT MASS / ATOMIC MASS RAW RATIO DIVIDE BY SMALLEST SUBSCRIPTRATIO Ag 85 g / 107.86g/mol =0.7880 / 0.7880 = 1.0 F 15 g / 18.99 g/mol =0.7898 Ag1F1 AgF Assume 100g of the sample, this will allow you to assume 85% is 85 grams. Total mass does NOT affect % composition.

NH3 (g) 2 X liters = = 20 N2 (g) 1 3 H2 (g) + N2 (g) 2 NH3 (g) #23) CALCULATE THE LITERS OF AMMONIA (NH3) GAS FORMED FROM 20 LITERS OF N2 GAS REACTED in the reaction given below. 3 H2 (g) + N2 (g) 2 NH3 (g) X = 40 Liters of ammonia gas NH3 (g) 2 X liters = = 20 N2 (g) 1 GAS VOLUMES CAN EXIST IN A RATIO AS DEFINED BY THE COEFFICIENTS OF THE BALANCED EQUATION, RATIO AS YOU WOULD RATIO MOLES! THIS ASSUMES P AND T ARE CONSTANT.

#28) WHAT IS THE MOLARITY OF A SOLUTION OF KNO3 (MOLECULAR MASS=101) THAT CONTAINS 404 GRAMS OF KNO3 IN 2.00 LITERS OF SOLUTION? MOLES = MASS GFM MOLARITY = MOLES VOLUME MOLES = 404. g = 4mol KNO3 101. g/mol MOLARITY = 4 mol = 2.00M 2.00L

X = 67.2 g/mol = 22.4 LITERS 1 LITER 3.00 g #15) The density of a gas is 3.00 g/L at STP, WHAT IS THE GFM OF THE GAS? 3.00 g X = 67.2 g/mol = 22.4 LITERS 1 LITER DENSITY RATIO, GIVEN 1 MOLE = 22.4L THE MASS OF 22.4 L IS THE MOLAR MASS AT STP