Mole
Conversions
Percent
Composition
Molecular and
Empirical
Formulas
Balancing
Equations
Stoichiometry
100
100
100
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200
200
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300
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500
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500
Mole Conversion for 100 points
¿How many moles are in 28
grams of CO2?
• 28.0gCO2 x 1 mol CO2 = 0.636 mol CO2
44.01gCO2
Mole Conversion for 200 points
How many moles of magnesium is 3.01 x
1022 atoms of magnesium?
• 3.01e22 atom Mg x 1 mole Mg = 0.0499 mol Mg
6.022e23 atoms Mg
Mole Conversion for 300 points
Determine the volume, in liters, occupied
by 0.030 moles of a gas at STP.
• 0.030 mol x 22.4 L = 0.67 L
1 mole
Mole Conversion for 400 points
How many oxygen molecules are
in 3.36 L of oxygen gas at STP?
• 3.36 L O2 x 1 mole O2 x 6.022e23 molecules = 9.03e22
22.4 L O2
1 mole O2
Mole Conversion for 500 point
Find the mass in grams of 2.00 x 1023
molecules of F2.
• 2.00e23 mc F2 x 1 mole F2 x 38g F2 = 12.6g F2
6.022e23 mc F2 1 mole
Percent Composition for 100 points
Find the percent composition of CuBr2
Cu =
63.55
Br x2 =+ 159.8
223.35
63.55(1) x 100 = 28.45%
223.35
79.9(2) x 100 = 71.55%
223.35
Percent Composition for 200 points
Find the percent composition of NaOH
Na = 22.99
O = 16.00
H = + 1.01
40.00
(1)22.99 x 100 = 57.48%
40.00
(1)16.00 x 100 = 40.00%
40.00
(1)1.01 x 100 = 2.53%
40.00
Percent Composition for 300 points
Find the percent composition of N2S2
N x2 = 28.02
S x2 =+64.02
92.04
14.01(2) x 100 = 30.44%
92.04
32.01(2) x 100 = 69.56%
92.04
Percent Composition for 400 points
Find the percent composition of
KMnO4
K = 39.10
Mn = 54.94
O x4= + 64.00
158.04
(1)39.10 x 100 = 24.74%
158.04
(1)54.94 x 100 = 34.76%
158.04
(4)16.00 x 100 = 40.50%
158.04
Percent Composition for 500 points
Find the percent composition of
Al2(SO4)3
Al x2= 53.96
S x3 = 96.03
O x12= + 192
341.99
(2)26.98 x 100 = 15.78%
341.99
(3)32.01 x 100 = 28.08%
341.99
(12)16.00 x 100 = 56.14%
341.99
Empirical and Molecular Formulas for 100 points
Find the empirical formula for a compound
containing .783% Carbon, .196% Hydrogen, and
.521% Oxygen.
.783g C x 1 mol C = 0.0652 = 2
12.01g C 0.0326
.196g H x 1 mol H = .194 = 6
1.01g H 0.0326
.521g O x 1 mol O = 0.0326 = 1
16.00g O 0.0326
=C2H6O
Empirical and Molecular Formulas for 200 points
The empirical formula for a substance is CH2O. If
its molar mass is 180, what is the molecular
formula
C
= 12.01
H x2 = 2.02
O = +16.00
30.03
180 = 6 = 6(CH2O) = C6H12O6
30.03
Empirical and Molecular Formulas for 300 points
The empirical formula for a substance is
C3H7NO3. If its molar mass is 105.11, what is the
molecular formula
C x3 = 36.03
H x7 = 7.07
N x1 = 14.01
O x3 = +48.00
105.11
105.11 = 1 = 1(C3H7NO3) = C3H7NO3
105.11
Empirical and Molecular Formulas for 400 points
Find the empirical formula for a compound containing
1.388g Carbon, .345g Hydrogen, and 1.850g Oxygen.
1.388g C x 1 mol C = 0.1156 = 1
12.01g C 0.1156
.345g H x 1 mol H = 0.3416 = 3
1.01g H 0.1156
1.850g O x 1 mol O = 0.1156 = 1
16.00g O 0.1156
=CH3O
Empirical and Molecular Formulas for 500 points
Find the empirical and molecular formula for a
compound containing 11.39g Phosphorus and 39.12g
Chloride. Its molar mass is 274.64g.
11.36g P x 1 mol P = 0.3677 = 1
30.97g P 0.3677
39.12g Cl x 1 mol Cl = 1.1035 = 3
35.45g Cl 0.3677
P x1 = 30.97
Cl x3 = +106.35
137.32
PCl3
274.64 = 2 = 2(PCl3) = P2Cl6
137.32
Balancing Equation for 100 points
Balance: _N2 + _H2  _NH3
Balanced: 1N2 + 3H2  2NH3
Balancing Equation for 200 points
Balance: _KClO3  _KCL + _O2
Balanced: 2KClO3  2KCL + 3O2
Balancing Equation for 300 points
Balance: _Na + _H2O  _NaOH + _H2
Balanced: 2Na + 2H2O  2NaOH + 1H2
Balancing Equation for 400 points
Balance: _FeCl3 + _NaOH  _Fe(OH)3 + _NaCl
Balanced: 1FeCl3 + 3NaOH  1Fe(OH)3 + 3NaCl
Balancing Equation for 500 points
Balance: _C8H18 + _O2  _CO2 + _H2O
Balanced: 2C8H18 + 25O2  16CO2 + 18H2O
Stoichiometry for 100 points
Given this equation:2 KClO3 ---> 2 KCl + 3 O2,
How many moles of O2 can be produced by
letting 12.00 moles of KClO3 react?
12 mol KClO3 x 3 mol O2 = 18 mol O2
2 mol KClO3
Stoichiometry for 200 points
Given this equation: 2K + Cl2 ---> 2KCl, how
many moles of KCl would be produced from
2.50g of K and an excess of Cl2
2.50g K x 1 mol K x 2 mol KCl = 0.0639 mol KCL
39.10g K 2 mol K
Stoichiometry for 300 points
Given this equation: 2NaClO3  2NaCl + 3O2,
how many grams of O2 would be produced from
12.0 moles of NaClO3
12 mol NaClO3 x 3 mol O2 x 32.0g O2 = 576g O2
2 mol NaClO3 1 mol O2
Stoichiometry for 400 points
Given this equation: Na2O + H2O ---> 2NaOH,
How many grams of NaOH can be produced
from 54.8 grams of Na2O and an excess of H2O
reacting
54.8g Na2O x 1 mol Na2O x 2 mol NaOH x 40 g NaOH = 35.4
61.98g Na2O 2 mol KClO3 1 mol NaOH
Stoichiometry for 500 points
Given this equation: 8Fe + S8 ---> 8FeS, How
many grams of FeS can be produced from 24.5
grams of S and an excess of Fe reacting
24.5g S8 x 1 mol S8 x 8 mol FeS x 87.86 g FeS = 67.2
256.08g S 1 mol S8 1 mol FeS