Problems 

Problem 3.1

•         1. What is the volume of Chlorine needed to produce 292.5 g of Sodium Chloride?

Na + Cl2 →   NaCl

Problem 3.2

•         Calculate atoms of sodium needed to produce 292.5 g of Sodium Chloride?

Problem 3.3

•         What is the mass of Sodium Chloride formed from 2.4088 X 1024 atoms Chlorine?

•         Na + Cl2 →   NaCl

Problem 3.1

•         1. What is the volume of Chlorine needed to produce 292.5 g of Sodium Chloride?

•         292.5 g /58.5g/mole = 5 moles NaCl

•         Set the ratio. 

•         2 Na + Cl2 →  2 NaCl

•                     X        5  X=2.5moles

•         Calculate volume of Chlorine

•          2.5 moles (22.4 L/mol)= 56 L

Problem 3.2

•         Convert to moles

•         Set the ratio. 

•         2 Na + Cl2 →  2 NaCl

•         X           5  X=5 moles

•         Calculate atoms of sodium

•          5moles / 6.022 X 1023 atoms/mol = 3.011 X 1024 atoms

Problem 3.3

•         What is the mass of Sodium Chloride formed from of 2.4088 X 1024 atoms Chlorine?

•         2.4088 X 1024 atoms / 6.022 X 1023 atoms/mol = 4 moles

•         Since Cl2 is diatomic 4 moles/2=2

•         Set the ratio. 

•         2 Na + Cl2 →  2 NaCl

•                     2         X  X=4 moles

•         Calculate mass of Sodium Chloride

•         Na=23 + Cl= 35.5 => NaCl=58.5 g/mol

•          4 moles (58.5 g/mol)= 234 g

Problem 7.1

•         When 3.6132 X 1024 molecules of Hydrogen Peroxide decompose, how many atoms of Oxygen would be formed?

•          H2O2 →  H2O + O2

Problem 7.2

•         What volume of Oxygen would be produced from 170 g of Hydrogen Peroxide?
H2O2 →  H2O +O2

Problem 7.3

•          67.2 L of Oxygen will react to produce what volume of water?

•          H2 + O2 →   H2O

Problem 7.1

•         When 3.6132 X 1024 molecules of Hydrogen Peroxide decomposes, how many atoms of Oxygen would be formed?

•         3.6132 X 1024 molecules/ 6.022 X 1023 molecules/mole =6 moles

•         Set the ratio. 

•         2 H2O2 → 2 H2O +1O2

•         6                                X X=3 moles

•         Calculate the mass of water

•          3 moles(6.022 X 1023 molecules /mol)*2 =3.6132 X 1024 atoms

Problem 7.2

•         What volume of Oxygen
would be produced from 170 g of Hydrogen Peroxide?
H2O2 →  H2O +O2

•         170 g /34 g/mol = 5 moles

•         Set the ratio. 

•         2 H2O2 → 1 O2   +2 H2O      

•         5           X           X=2.5moles

•         Calculate the volume of oxygen

•          2.5moles (22.4 L/mol)= 56 L

Problem 7.3

•          67.2 L of Oxygen will react to produce how what volume of water?

•         67.2/22.4=3moles

•         Set the ratio. 

•         2 H2 +1 O2 →  2 H2O

•                          3                  X   X= 6 moles

•         Calculate the volume of water

•          6 X 22.4= 134.4 L

Problem 6.1

•         What is the volume of Chlorine consumed 135 g by Aluminum?
  Al + Cl2 →   AlCl3

Problem 6.2

•         2.How many formula units of Aluminum Chloride are formed from 135 g of Aluminum?

•         Al + Cl2 →   AlCl3

Problem 6.3

•         3.How many Chlorine atoms react with 135 g of Aluminum?

•         Al + Cl2 →   AlCl3

Problem 6.4

•         4. How many Aluminum atoms are consumed?
  Al + Cl2 →   AlCl3

Problem 6.1

•         What is the volume of Chlorine consumed 135 g by Aluminum?
  2Al +3 Cl2 →  2 AlCl3
135 g /27g/mole = 5 moles Al

•         Set the ratio. 

     2Al +3 Cl2 →  2 AlCl3
      5       X           X=7.5moles

•         Calculate volume of Chlorine

•          7.5 moles (22.4 L/mol)= 168 L

Problem 6.2

•         2.How many formula units of Aluminum Chloride are formed in the reaction?

•         2Al +3 Cl2 →  2 AlCl3
135 g /27g/mole = 5 moles Al

•         Set the ratio. 

     2Al +3 Cl2 →  2 AlCl3
      5                X   X=5 moles

•          5moles*6.022 X 1023 atoms/mol = 3.011 X 1024 atoms

Problem 6.3

•         3.How many Chlorine atoms are consumed?

•         2Al +3 Cl2 →  2 AlCl3
135 g /27g/mole = 5 moles Al

•         Set the ratio. 

     2Al +3 Cl2 →  2 AlCl3
      5        X            X=7.5 moles

•          7.5moles*6.022 X 1023 molecules/mol = 45.165 X 1023 molecules

•         45.165 X 1023 molecules*2=9.033 X 1024 atoms

Problem 6.4

•         4. How many Aluminum atoms are consumed?
  2Al +3 Cl2 →  2 AlCl3
135 g /27g/mole = 5 moles Al

•         Calculate

•         5moles Al*6.022 X 1023 atoms/mol = 3.011 X 1024 atoms

Problem 5.1

1.  What is the volume of Nitrogen consumed by 160 g of Calcium?
     Ca + N2 →  Ca3N2

•         2.How many atoms of Calcium are consumed in the reaction?
  Ca + N2 →  Ca3N2

•         3. What is the mass of Calcium Nitride formed in the reaction?

•          Ca + N2 →  Ca3N2

•         What is the volume of Nitrogen consumed by 160 g of Calcium?
3 Ca + N2 →  Ca3N2

•         160 g /40g/mole = 4 moles Ca

•         Set the ratio. 

•         3 Ca + 1 N2 →  Ca3N2
4            X               X=1.3 moles

•         Calculate volume of Chlorine

•          1.3 moles (22.4 L/mol)= 29.12 L

Problem 5.2

•         2.How many atoms of Calcium are consumed in the reaction?
3 Ca + N2 →  Ca3N2

•         160 g /40g/mole = 4 moles Ca

•         4 moles (6.022 X 1023)=

•         24.088 X 1023 atoms  or

•         2.4088 X 1024 atoms

Problem 5.3

•         3. What is the mass of Calcium Nitride formed in the reaction?

•         3 Ca + N2 →  Ca3N2

•         160 g /40g/mole = 4 moles Ca

•         Set the ratio. 

•         3 Ca + N2 →  1 Ca3N2
4                            X            X=1.3 moles

•         Calculate mass of Ca3N2

•          1.3 moles (148g/mol)= 197.3 g

Problem 4.1

•         What is the volume of Nitrogen consumed by 160 g of Calcium?
3 Ca + N2 →  Ca3N2

•         160 g /40g/mole = 4 moles Ca

•         Set the ratio. 

•         3 Ca + 1 N2 →  Ca3N2
4            X               X=1.3 moles

•         Calculate volume of Chlorine

•          1.3 moles (22.4 L/mol)= 29.12 L

Problem 4.2

•         2.How many atoms of Calcium are consumed in the reaction?
3 Ca + N2 →  Ca3N2

•         160 g /40g/mole = 4 moles Ca

•         4 moles (6.022 X 1023)=

•         24.088 X 1023 atoms  or 2.4088 X 1024 atoms

Problem 4.3

•         3. What is the mass of Calcium Nitride formed in the reaction?

•         3 Ca + N2 →  Ca3N2

•         160 g /40g/mole = 4 moles Ca

•         Set the ratio. 

•         3 Ca + N2 → 1 Ca3N2
4                            X            X=1.3 moles

•         Calculate mass of Ca3N2

•          1.3 moles (148g/mol)= 197.3 g

Problem 3.1

•         1. What is the volume of Chlorine needed to produce 292.5 g of Sodium Chloride?

•         292.5 g /58.5g/mole = 5 moles NaCl

•         Set the ratio. 

•         2 Na + Cl2 →  2 NaCl

•                     X        5  X=2.5moles

•         Calculate volume of Chlorine

•          2.5 moles (22.4 L/mol)= 56 L

Problem 3.2

•         Convert to moles

•         Set the ratio. 

•         2 Na + Cl2 →  2 NaCl

•         X           5  X=5 moles

•         Calculate atoms of sodium

•          5moles / 6.022 X 1023 atoms/mol = 3.011 X 1024 atoms

Problem 3.3

•         What is the mass of Sodium Chloride formed from of 2.4088 X 1024 atoms Chlorine?

•         2.4088 X 1024 atoms / 6.022 X 1023 atoms/mol = 4 moles

•         Since Cl2 is diatomic 4 moles/2=2

•         Set the ratio. 

•         2 Na + Cl2 →  2 NaCl

•                     2         X  X=4 moles

•         Calculate mass of Sodium Chloride

•         Na=23 + Cl= 35.5 => NaCl=58.5 g/mol

•          4 moles (58.5 g/mol)= 234 g

Problem 2.1

•         When 1.806 X 1024 molecules of Hydrogen Peroxide decomposes, how many grams of water would formed?

•         1.806 X 1024 molecules/ 6.02 X 1023 molecules/mole =3 moles

•         Set the ratio. 

•         2 H2O2 →  2 H2O +O2

•         3      X         X=3 moles

•         Calculate the mass of water

•          3moles (18g/mol)= 54 g

Problem 2.2

•         B) When 1.806 X 1024 molecules of Hydrogen Peroxide decomposes, what volume of Oxygen
would be produced?

•         1.806 X 1024 molecules/ 6.02 X 1023 molecules/mole =3 moles

•         Set the ratio. 

•         2 H2O2 →  1 O2   +2 H2O      

•         3           X           X=1.5moles

•         Calculate the volume of oxygen

•         1.5moles (22.4 L/mol)= 33.6L

Problem 2.3

•         2. 67.2 L of Oxygen will react with how many molecules of Hydrogen to produce water?

•         67.2/22.4=3moles

•         Set the ratio. 

•         2 H2 + O2 →  2 H2O

•         X         3  X= 6 moles

•         Calculate the number of molecules

•          6 moles(6.02 X 1023 molecules/mol) =3.612 X 1024 molecules

Problem 1.1

•         1.What is the volume of Chlorine
produced as 1.2044 X 1024 Formula units of Aluminum Chloride Decompose?

•         1.2044 X 1024 Form. units/ 6.022 X 1023 Form. units/mole = 2 moles AlCl3

•         Set the ratio. 

•         2 Al + 3 Cl2 ←   2 AlCl3

•                    X          2 X=3 moles

•         Calculate volume of Chlorine

•          3 moles (22.4 L/mol)= 67.2 L

Problem 1.2

•         2.How many atoms of Aluminum are formed in the reaction?

•         1.2044 X 1024 Form. units/ 6.022 X 1023 Form. units/mole = 2 moles AlCl3

•         Set the ratio. 

•         2 Al + 3 Cl2 ←   2 AlCl3

•         X                        2 X=2 moles

•         Calculate atoms of Aluminum

•          2 moles (6.022 X 1023 atoms/mol)= 1.2044 X 1024 atoms of Aluminum

Problem 1.3

•         3. What mass of Aluminum Chloride needed to decompose into 710 g of Chlorine?

•         710 g/ 71 g/mole = 10 moles Cl2

•         Set the ratio. 

•         2 Al + 3 Cl2 ←  2 AlCl3    --

•                   10        X X=6.6 moles

•         Calculate mass of Aluminum Chloride

•          6.6 moles (133.5 g/mol)= 881.1 g

CONVERSION PROBLEMS

Problem 1.1

•         Given 452g of Radium, how many atoms are present?

•         Convert to moles –P. table.

•         452 g/ (226 g/mol) =2 moles

•         2 moles( 6.022 X 1023 atoms/mol)

•         12.044 X 1023atoms

•         1.2044 X 1024atoms

Problem 1.2

•         What is the mass of Bromine if you have 44.8 L?

•         Convert to moles-vol 22.4

•         44.8 L/(22.4 L/mol)=2 moles

•         Calculate the mass-P. table

•         80 g/mol (2-diatomic) =160 g/mol

•          2 moles(160g/mol)

•         320 g

Problem 1.3

•         What is the mass of tin if you are given 1.2044 X 1024 atoms of tin?

•         1.2044 X 1024 / (6.022 X 1023 atoms/moles)

•         2 moles of tin. 

•         Calculate mass of Tin.

•          2 moles (119 g/mol)= 238 g

Problem 2.1

•         Given 28 g of Lithium, how many atoms are present?
Convert to moles –P. table.

•         28 g/ 7 g/mol =4 moles

•         4 moles( 6.022 X 1023 atoms/mol)

•         24.088 X 1023 atoms

•         2.4088 X 1024 atoms

Problem 2.2

•         What is the mass of Helium if you have 224 L?

•         Convert to moles-vol 22.4

•         224 L/22.4 L/mol=10 moles

•         Calculate the mass-P. table

•          10 moles(4 g/mol)

•         40 g

Problem 2.3

•         What is the mass of Antimony if you are given 1.2044 X 1024 atoms of Antimony?

•         1.2044 X 1024 / 6.022 X 1023 atoms/moles

•         2 moles of Sb. 

•         Calculate mass of Sb.

•         Mass from periodic table is 122

•          2 moles (122 g/mol)= 244 g

MIXED PROBLEMS

Problem 6.1

•         Given 208 g of Chromium, how many atoms are present?
Convert to moles –P. table.

•         208 g/ 52 g/mol =4 moles

•         4 moles( 6.02 X 1023 atoms/mol)

•         24.08 X 1023 atoms

•         2.408 X 1024atoms

Problem 6.2

•         What is the mass of Iodine if you have 11.2 L?
Convert to moles-vol 11.2

•         11.2 L/22.4 L/mol=.5 moles

•         Calculate the mass-P. table

•         Iodine = 127g/mole  I2(254)

•          .5 moles(254 g/mol)

•         127 g

Problem 6.3

•         Calculate % mass for C6F4Cl2Si3

•         Atom   mass #

•         C                 12                6= 72

•         F                  19                4=76

•         Cl                35.5             2= 71

•         Si                 28                3=84

•         Total mass= 72+76+71+84=303 g/mol

•         C=72/303  F=76/303  Cl= 71/303 Si=84/303

•         C=23.7%  F=25%  Cl= 23.4%  Si=27.7%

Problem 6.4

Write the proper formula if the mass is 387 grams and the empirical formula is P3H4O2 387/129=3
Multiply each subscript by 3.
P9H12O6

Problem 5.1

•         Calculate the mass for 1.8066 X 1024 molecules of Bromine.
Convert to moles –Av #.

•         1.8066 X 1024 /( 6.022 X 1023 molecules/mol)= 3 moles

•         3 moles( 80g/mol)=240 g

•         240 g (2 because it is diatomic)= 480 g

Problem 5.2

•         Given 89.6 L  of Krypton, calculate the mass and number of atoms.
Convert to moles-vol 22.4

•         89.6 L/22.4 L/mol=4 moles

•         Calculate the mass-P. table

•          4 moles(84 g/mol)=336 g

•         Calculate the # of atoms.

•         4 moles ( 6.022 X 1023 atoms/mol)= 2.4088 X 1024 atoms

Problem 5.3

3. Calculate % mass for S3P4BrCl2
Atom   mass #

•         S                  32                3= 96

•         P                  31                4= 124

•         Br                80                1=80

•         Cl                35.5             2 = 71

•         Total mass= 96+124+80+71=371 g/mol

•         S=32/371  P=124/371  Br= 80/371

•         Cl=71/371

•         S=26%  P=34%   Br= 21.5%  Cl=19.14%

Problem 5.4

Write the proper empirical formula for C4N2O7 and Cl3P6O9
Find the least common multiple and divide.

  (NONE)    C4N2O7
 Divide by 3   ClP2O3