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MOLES REVIEW, MAY 28

MOLES REVIEW, MAY 28. 2 Al 2 O 3  3 O 2 + 4 Al a) What is the mole ratio of O 2 to Al?. O 2 = Al. Al 2 O 3  O 2 + Al O 2 = O 2 = Al = Al =. 2. 3. 2. 4. 6. 6. 4. 2. 4. 2.

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MOLES REVIEW, MAY 28

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

  2. 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

  3. 2 Al2O3 3 O2 + 4 Al d) What is the % by mass composition of Al2O3 ? NEVER USE COEFFICENTS FOR MOLAR MASS CALCULATIONS % COMP (MASS) = PART X 100 WHOLE 101.93 g/mol First you need to find the molar mass (gfm) of the substance

  4. % COMP (MASS) = PART X 100 WHOLE % COMP Al = 53.96 g X 100 = 52.93 % 101.93 g 101.93 g/mol

  5. 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. 13.017 g/mol THIS IS THE MASS (GFM)OF THE EMPIRICAL FORMULA

  6. 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

  7. 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. Ag1F1 AgF Assume 100g of the sample, this will allow you to assume 85% is 85 grams. Total mass does NOT affect % composition.

  8. #23) CALCULATE THE LITERS OF AMMONIA (NH3) GAS FORMED FROM 20 LITERS OF N2GAS 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.

  9. #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 = 4molKNO3 101. g/mol MOLARITY = 4 mol =2.00M 2.00L

  10. #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

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