# Review - Solution Stoichiometry and the Mole

Chem I | Chem II AP
1) Given: Cr-52 # of p+ = # of e- = # of no = mass # = Atomic mass = u = atom (s) g-atomic mass = g = atom (s) Protons and electrons are equal to the atomic number.Neutrons are equal to the mass number minus the protons.The atomic mass is the mass of one atom in u, while the g-atomic mass is the mass of a mole of atoms in grams.
2) What is the mass of: one 31P atom = u one mole of 31P atoms = g 6.02 x 1023 atoms of 31P = g Use 31 instead of 30.97 because 31P is referring to a specific isotope of phosphorus that has a mass of 31.
3) Given: NO2 Formula weight (mass) = u g-formula weight = g g-molecular weight = g
4) Mass of one mole of average Mg atoms = g Mass of one mole of 24Mg atoms = g The average mass of an element is given on the periodic table. 24Mg refers to a specific isotope of Mg that has a mass of 24.
5)
49.0 g of H2SO4 =
moles
49.0 g H2SO4 x
 1 mol 98.1 g
= 0.5 mol
6)
0.10 mole of carbon =
atoms
0.10 mol C x
 6.02 x 1023 atoms 1 mol
= 6.02 x 1022 atoms
7)
2.0 x 10-2 moles of CO2 =
g
2.0 x 10-2 mol CO2 x
 44 g 1 mol
= 0.88 g
8)
26.0 g of H2O =
molecules
26.0 g H2O x
 6.02 x 1023 molecules 18.0 g
= 8.70 x 1023 molecules
9)
26.0 g of H2O =
total atoms
26.0 g H2O x
 6.02 x 1023 molecules 18.0 g
x
 3 atoms 1 molecule
= 2.61 x 1024 total atoms
10)
% oxygen in Pb(NO3)2 =
%
 96 g 331.2 g
=
 x 100
How many g of Pb(NO3)2 contain 56.0 g of nitrogen =
g
 28 g 331.2 g
=
 56 g x
11)
# of g of oxygen in 50.0 g of MgCO3 =
g
 48 g 84.3 g
=
 x 50.0 g
12)
Determine the empirical formula for the following compound:
40.3% K; 26.7% Cr; 33.0% O
40.3 % K = 40.3 g K x
 1 mol 39.1 g
= 1.03 mol / 0.513 = 2
26.7 % Cr = 26.7 g x
 1 mol 52.0 g
= 0.513 mol / 0.513 = 1
33 % O = 33 g x
 1 mol 16 g
= 2.06 / 0.513 = 4

K2CrO4

13)
Determine the molecular formula for the following compound:
68.8% C; 4.92% H; and oxygen
g-MW = 122 g/mol
68.8 % C = 68.8 g x
 1 mol 12 g
= 5.73 mol / 1.64 = 3.5 x 2 = 7
4.92 % H = 4.92 g x
 1 mol 1 g
= 4.92 mol / 1.64 = 3 x 2 = 6
26.28 % O = 26.28 g x
 1 mol 16 g
= 1.64 mol / 1.64 = 1 x 2 = 2
 C7H6O2 has a molecular mass of 122 g, so the empirical and molecular formulas are the same.
14)
Determine the mass of NaOH (solid) needed to prepare 1.00 liter of 3.0 M NaOH solution:
g
3M =
 x 1 L
x = 3 mol x
 40 g 1 mol
= 120 g
 Describe the preperation of the solution:
15)
Determine the mass of CaCl2 (solid) needed to prepare 250.0 mL of 0.50 M CaCl2 solution:
g
0.50 M =
 x 0.25 L
x = 0.125 mol x
 111.1 g 1 mol
= 13.9 g
 Describe the preperation of the solution:
16)
Determine the volume of 6.0 M HCl needed to make 2.0 liter of 3.0 M HCl:
L
 M1V1 = M2V2
 (3.0 M)(2.0 L) = (6.0 M)(x) x = 1.0 L
 Describe the preperation of the solution:
17)
How many grams of magnesium are required to react with excess oxygen to produce 2.00 g of magnesium oxide?
g
 2 Mg ? g
 +
 O2
 →
 2 MgO 2.00 g
2.00 g MgO x
 1 mol MgO 40.3 g MgO
x
 2 mol Mg 2 mol MgO
x
 24.3 g Mg 1 mol Mg
= 1.21 g Mg
18)
How many grams of copper will be produced from the reaction of 3.00 g of zinc with 5.00 g copper (II) nitrate?
g
 Zn 3.00 g
 +
 Cu(NO3)2 5.00 g
 →
 Cu ? g
 +
 Zn(NO3)2
3.00 g Zn x
 1 mol Zn 65.4 g Zn
x
 1 mol Cu 1 mol Zn
x
 63.5 g Cu 1 mol Cu
= 2.91 g Cu
5.00 g Cu(NO3)2 x
 1 mol Cu(NO3)2 187.5 g Cu(NO3)2
x
 1 mol Cu 1 mol Cu(NO3)2
x
 63.5 g Cu 1 mol Cu
= 1.69 g Cu
How many grams of the excess reactant are left over?
g
5.00 g Cu(NO3)2 x
 1 mol Cu(NO3)2 187.5 g Cu(NO3)2
x
 1 mol Zn 1 mol Cu(NO3)2
x
 65.4 g Zn 1 mol Zn
= 1.74 g Zn used
 3.00 g Zn initially - 1.74 g Zn used = 1.26 g Zn left over
19)
How many grams of magnesium are required to react with 100.0 mL of 0.50 M nitric acid solution?
g
 Mg + 2 HNO3 → Mg(NO3)2 + H2
0.50 M =
 x 0.100 mL
x = 0.050 mol HNO3
0.050 mol HNO3 x
 1 mol Mg 2 mol HNO3
x
 24.3 g Mg 1 mol Mg
= 0.61 g Mg
20)
How many grams of PbI2 precipitate will be formed from the reaction of 10.0 mL of 0.50 M Pb(NO3)2 with 10.0 mL of 0.50 M NaI?
g
 Pb(NO3)2 + 2 NaI → PbI2 + 2 NaNO3
0.50 M =
 x 0.010 L
x = 0.005 mol
0.005 mol Pb(NO3)2 x
 1 mol PbI2 1 mol Pb(NO3)2
x
 461 g PbI2 1 mol PbI2
= 2.31 g PbI2
0.005 mol NaI x
 1 mol PbI2 2 mol NaI
x
 461 g PbI2 1 mol PbI2
= 1.15 g PbI2
21)
How many grams of hydrogen will be produced from the reaction of 10.0 g of magnesium and 50.0 mL of 0.50 M HCl soution?
g
 Mg + 2 HCl → H2 + MgCl2
0.50 M =
 x 0.050 L
x = 0.025 mol HCl x
 1 mol H2 2 mol HCl
x
 2 g H2 1 mol H2
= 0.025 g
10 g Mg x
 1 mol Mg 24.3 g Mg
x
 1 mol H2 1 mol Mg
x
 2 g H2 1 mol H2
= 0.823 g
22)
Determine the % yield if 10.0 g of H2 react with excess O2 to produce 80.0 g of water
%
 2 H2 + O2 → 2 H2O
10.0 g H2 x
 1 mol H2 2 g H2
x
 2 mol H2O 2 mol H2
x
 18 g H2O 1 mol H2O
= 90 g H2O
 80 g 90 g
=
 x 100
x = 88.9%

Theoretically, 90 g of H2O should have been produced, but only 80 g were produced. This corresponds to an 88.9% yield.

23)
Given:
TiO2 + 2 Cl2 + 2 C → TiCl4 + 2 CO
TiCl4 + 2 Mg → Ti + 2 MgCl2
How many grams of titanium can be produced from the processing of 4500 g of ore that is 50.0% Carbon? [Assume sufficient TiO2 and Cl2]
g
 TiO2 + 2 Cl2 + 2 C → TiCl4 + 2 CO
 TiCl4 + 2 Mg → Ti + 2 MgCl2
 x 4500 g
=
 50 100
x = 2250 g Of the 4500 g of ore, 2250 g is Carbon.
2250 g C x
 1 mol C 12 g C
x
 1 mol TiCl4 2 mol C
= 93.75 mol TiCl4
93.75 mol TiCl4 x
 1 mol Ti 1 mol TiCl4
x
 47.9 g Ti 1 mol Ti
= 4490 g Ti
24)
1.55 g of CHO compound are combusted for analysis which yields 1.45 g of CO2 and 0.89 g of water. Find the empirical formula.
 C?H?O? 1.55 g
 +
 O2
 →
 CO2 1.45 g
 +
 H2O 0.8 g
C:
 12 g 44 g
=
 x 1.45 g
x = 0.395 g C x
 1 mol 12 g
= 0.0329 mol / 0.0329 mol = 1
H:
 2 g 18 g
=
 x 0.89 g
x = 0.0989 g H x
 1 mol 1 g
= 0.0989 mol / 0.0329 mol = 3
O:1.55 g - 0.395 g - 0.0989 g = 1.056 g O x
 1 mol 16 g
= 0.0660 mol / 0.0329 mol = 2