Download + O 2

Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 3
Stoichiometry: Calculations
with Chemical Formulas
and Equations
Troy Wood
University of Buffalo
Buffalo, NY
 2006, Prentice Hall
The following equation is unbalanced:
CS2 + O2
CO2 + SO2
What is the correct balanced equation?
1.
2.
3.
4.
CS2 + 2 O2
2 CS2 + 12 O2
CS2 + 3 O2
CS2 + 6 O2
CO2 + 2 SO2
2 CO2 + 4 SO2
CO2 + 2 SO2
CO2 + 2 SO2
Correct Answer:
1.
2.
3.
4.
CS2 + 2 O2
2 CS2 + 12 O2
CS2 + 3 O2
CS2 + 6 O2
CO2 + 2 SO2
2 CO2 + 4 SO2
CO2 + 2 SO2
CO2 + 2 SO2
This is the only case where the number of
each type of atom is the same on both
the reactant and product sides of the
equation.
Which of the following is the correct,
balanced chemical equation for the
combustion of ethanol (C2H5OH)?
1.
2.
3.
4.
C2H5OH + O2
C2H5OH + 3 O2
C2H5OH + 6 O2
2 C2H5OH + 7 O2
CO2 + H2O
2 CO2 + 3 H2O
4 CO2 + 6 H2O
4 CO2 + 6 H2O
O
O
C
C
O
Correct Answer:
1.
2.
3.
4.
C2H5OH + O2
C2H5OH + 3 O2
C2H5OH + 6 O2
2 C2H5OH + 7 O2
CO2 + H2O
2 CO2 + 3 H2O
4 CO2 + 6 H2O
4 CO2 + 6 H2O
C
OO
O
C
O
If the empirical formula of a compound is
C2H2N but the molecular weight is
160 g/mol, what is the molecular
formula?
1.
2.
3.
4.
5.
C2H2N
C4H4N2
C6H8N3
C8H8N4
None of the above choices is correct.
Correct Answer:
1.
2.
3.
4.
5.
C2H2N
C4H4N2
C6H8N3
C8H8N4
None of the above choices is correct.
FW C2H2N = (24 + 2 + 14) g/mol = 40 g/mol
MW/FW = 160/40 = 4
Thus, molecular formula is 4 empirical formula
4(C2H2N) = C8H8N4
How many fluorine atoms are present in
2.0 moles of AlF3?
1.
2.
3.
4.
5.
3.0  1023
6.0  1023
1.2  1024
3.6  1024
None of the above answers is correct.
Correct Answer:
1.
2.
3.
4.
5.
3.0  1023
6.0  1023
1.2  1024
3.6  1024
None of the above answers is correct.
(2.0 moles AlF3)(6.0  1023 molecules AlF3/mole AlF3)
(3 atoms F/1 molecule AlF3) = 3.6  1024
Which of the following samples has
the largest number of atoms?
1.
2.
3.
4.
5.
28 g Si
28 g CO
28 g N2
15 g C2H4
All of the above
have equal numbers
of atoms.
Correct Answer:
1.
2.
3.
4.
5.
28 g Si
28 g CO
28 g N2
15 g C2H4
All of the above
have equal numbers
of atoms.
28 g Si has 1 mole of atoms,
whereas 28 g of CO and N2
have 2 moles of atoms, even
though they also have only 1
mole substance. However,
there are 6 atoms for every 1
mole of C2H4, so in this case
there are slightly more than
3 moles worth of atoms.
The molecular mass for the compound,
XClO3, is 122.5 amu. The atomic mass of
X is equal to ____ amu, which is the
element ____.
1.
2.
3.
4.
5.
1.0; H
6.9; Li
23.0; Na
39.1; K
85.5; Rb
Correct Answer:
1.
2.
3.
4.
5.
1.0; H
6.9; Li
23.0; Na
39.1; K
85.5; Rb
MW = 122.5 amu = AW X + AW Cl + 3(AW O)
122.5 amu  AW Cl  3 (AW O) = AW X
122.5 amu  (35.45 amu)  3 (16.0 amu) = AW X
122.5  35.45 amu  48.0 amu = AW X
AW X = 39.1 amu
When the chemical equation,
C4H10(g) + O2(g)
H2O(g) + CO2(g),
is balanced, the stoichiometric coefficients
are, respectively:
1.
2.
3.
4.
1, 4, 5, 4
2, 13, 10, 8
2, 9, 8, 10
2, 7, 4, 5
Correct Answer:
1.
2.
3.
4.
1, 4, 5, 4
2, 13, 10, 8
2, 9, 8, 10
2, 7, 4, 5
The combustion of methane (CH4) is shown
below:
CH4(g) + 2 O2(g)
2 H2O(l) + CO2(g)
How many grams of water are produced in the
combustion of 1.0 moles of methane?
1.
2.
3.
4.
16 g
32 g
36 g
48 g
Correct Answer:
1.
2.
3.
4.
16 g
32 g
36 g
48 g
(1.0 mole CH4)(2 moles H2O/1 mole CH4)
(18 g H2O/1 mole H2O) = 36 g
0.5 mol of carbon is combusted with 1.0 mol of
oxygen gas. Which of the following statements
is correct?
1. There is a molar excess of carbon, not all carbon is
consumed.
2. There is a molar excess of oxygen, not all oxygen is
consumed.
3. There are stoichiometrically equivalent amounts of
carbon and oxygen gas.
Correct Answer:
1. There is a molar excess of carbon, not all carbon is
consumed.
2. There is a molar excess of oxygen, not all oxygen is
consumed.
3. There are stoichiometrically equivalent amounts of
carbon and oxygen gas.
C and O2 react in a 1:1 stoichiometric ratio,
therefore there must be excess O2 because there
was a larger initial number of moles of it.
0.5 mol of carbon is combusted with
1.0 mol g of oxygen gas. How many
moles of carbon dioxide are produced?
1.
2.
3.
4.
1.5 mol
1.0 mol
0.5 mol
2.0 mol
Correct Answer:
1.
2.
3.
4.
1.5 mol
1.0 mol
0.5 mol
2.0 mol
C and O2 react in a 1:1 stoichiometric ratio,
therefore C must be the limiting reagent and O2
must be the excess reagent. Initially,
(0.5 moles C)(1 mole CO2/1 mole C) = 0.5 mole CO2
24.0 g of carbon are combusted with
32.0 g of oxygen gas. How many grams
of carbon dioxide are produced?
1.
2.
3.
4.
56.0 g
88.0 g
28.0 g
44.0 g
Correct Answer:
1.
2.
3.
4.
56.0 g
88.0 g
28.0 g
44.0 g
C and O2 react in a 1:1 stoichiometric ratio. Initially,
(24.0 g C)  (1 mole C/12.0 g C)  (1 mole CO2/1 mole C) =
2.0 mole CO2
(32.0 g O2)  (1 mole O2/32.0 g O2)  (1 mole CO2/1 mole O2)
= 1.0 mole CO2
Thus, O2 is limiting, and 1.0 mole of CO2 is 44.0 g.
3.0 moles Mg are combusted with 2.0
moles O2. What is the limiting reagent
and how many moles of the excess
reagent remain?
1.
2.
3.
4.
Magnesium, 0.5 mole
Magnesium, 1.0 mole
Oxygen gas, 0.5 mole
Oxygen gas, 1.0 mole
Correct Answer:
1.
2.
3.
4.
Magnesium, 0.5 mole
Magnesium, 1.0 mole
Oxygen gas, 0.5 mole
Oxygen gas, 1.0 mole
Mg and O2 react in a 2:1 stoichiometric ratio.
Initially,
(3 mole Mg)(2 mole MgO/2 mole Mg)
= 1.5 mole MgO
(2 mole O2)(2 mole MgO/1 mole O2) = 4 mole MgO
Thus, Mg is limiting, and 3 mol Mg will consume
1.5 mol of O2, leaving 0.5 mole of O2.