1. 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. 2 Al2O3  3 O 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 O 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. 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 +
g/mol
First you need to find the molar mass (gfm) of the substance

4. 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 = g X 100 = % 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
g/mol

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

6. REVIEW CLASS MULTIPLE CHOICE QUESTIONS,
FRIDAY JUNE
GIVEN IN PROBLEM
#7)CONTINUED
MOLAR MASS
78. g/mol
5.99 = 6 = =
EMPIRICALMASS
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
#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
/ g/mol
=0.7880 /
= 1.0 F
15 g
/ 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.

8. 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.

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 = g = 4mol KNO3
101. g/mol
MOLARITY = 4 mol = 2.00M
2.00L

10. 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