CHE141
Chapter 3
Chapter 3
Stoichiometry: Calculations with Chemical Formulas and Equations
1. When the following equation is balanced, the coefficients are ____________.
NH3 (g) + O2 (g)
NO2 (g) + H2O (g)
(a). 1, 1, 1, 1
(b). 2, 3, 2, 3
(c). 4, 7, 4, 6
(d). 1, 3, 1, 2
Explanation: When dealing with any equation with molecular O2 avoid using
fractional coefficients.
2. When the following equation is balanced, the coefficients are ____________.
Al(NO3)3 + Na2S
Al2S3 + NaNO3
(a). 4, 6, 3, 2
(b). 2, 1, 3, 2
(c). 1, 1, 1, 1
(d).2, 3, 1, 6
3. When the following equation is balanced, the coefficient of Al2O3 is _________.
Al2O3 (s) + C (s) + Cl2 (g)
AlCl3 (s) + CO (g)
(a) 4
(b). 3
(c). 2
(d).1
4. When the following equation is balanced, the coefficient of H2S is____________.
FeCl3 (aq) + H2S (g)
Fe2S3 (s) + HCl (aq)
(a). 1
(b).3
(c). 2
(d). 5
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CHE141
Chapter 3
5. When the following equation is balanced, the coefficient of HCl is __________.
CaCO3 (s) + HCl (aq)
CaCl2 (aq) + CO2 (g) + H2O (1)
(a). 0.5
(b).2
(c). 3
(d). 4
6. When the following equation is balanced, the coefficient of C3H8O3 is ___________.
C3H8O3 (g) + O2 (g)
CO2 (g) + H2O (g)
(a). 4
(b). 3
(c). 2
(d). 7
7. When the following equation is balanced, the coefficient of O2 is__________.
C2H4O (g) + O2 (g)
CO2 (g) + H2O (g)
(a). 5
(b). 3
(c). 4
(d). 2
8. When the following equation is balanced, the coefficient of hydrogen is __________.
K (s) + H2O (l)
KOH (aq) + H2 (g)
(a). 2
(b).1
(c). 3
(d). 4
9. When the following equation is balanced, the coefficient of dinitrogen pentoxide is
______________.
N2O5 (g) + H2O (l)
HNO3 (aq)
(a). 4
(b). 2
(c). 3
(d).1
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CHE141
Chapter 3
10. When the following equation is balanced, the coefficient of nitric acid is __________.
(a). 5
(b).2
(c). 3
(d). 4
N2O5 (g) + H2O (l)
HNO3 (aq)
11. The balanced equation for the decomposition of sodium azide is __________.
(a). 2NaN3 (s)
(b). 2NaN3 (s)
(c). NaN3 (s)
(d). None of the above
2 Na (s) + 3N2 (g)
Na2 (s) + 3N2 (g)
Na (s) + N2 (g)
12. How many moles of carbon atoms are in 4 mol of dimethylsulfoxide (C2H6SO)?
(a). 2
(b). 6
(c). 8
(d). 4
Explanation: This is based on reading the formula and correctly extracting
information from it. The formula C2H6SO indicates that every mole of this
compound has 2 moles of carbon atoms in it. Thus 4 moles of the compound would
have 4 x 2 = 8 moles of C atoms.
13. There are ______ sulfur atoms in 25 molecules of C4H4S2.
(a). 1.5 x 1025
(b). 4.8 x 1025
(c). 3.0 x 1023
(d).50
Explanation: The molecular formula indicates that every molecule of C4H4S2 has 2
sulfur atoms per molecule and hence 25 molecules of this compound will have 25 x 2
= 50 atoms of sulfur.
14. There are _______ hydrogen atoms in 25 molecules of C4H4S2.
(a). 25
(b). 3.8 x 1024
(c). 6.0 x 1025
(d).100
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CHE141
Chapter 3
Explanation: The formula of C4H4S2 indicates that there are 4 hydrogen atoms per
molecule and hence 100 hydrogen atoms in 25 molecules of C4H4S2.
15. How many carbon atoms are contained in a sample of C3H8O that contains 200
molecules?
(a). 600
(b). 200
(c). 3.61 x 1026
(d). 1.20 x 1026
16. How many grams of oxygen are in 65.0 g of C2H2O2?
(a). 18
(b). 29
(c). 9.5
(d).35.8
Explanation: This question uses the mole to mole ratio between oxygen and C2H2O2
and needs the following steps.
65.0 g C 2H 2O 2
2 moles O
15.99 g O
$
$
= 35.8 g of O
#1
58.0 g " mol
1 mole C 2H 2O 2 1 mole of O
17. How many moles of carbon dioxide are there in 52.06 g of carbon dioxide?
!
(a). 0.8452
(b).1.183
(c). 1.183 x 1023
(d). 8.648 x 102
Explanation: This is a straight-forward conversion from grams to moles of CO2
which is done as follows:
1 mole CO 2
52.06 g CO 2 !
= 1.183 moles of CO 2
43.99 g CO 2
18. How many moles of the compound magnesium nitrate, Mg(NO3)2, are in a2.35 g
sample of this compound?
(a)
(b)
(c)
(d)
38.4
65.8
0.0158
0.0261
Explanation: This is a straight-forward conversion from grams to moles of
Mg(NO3)2 which is done as follows:
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 3
2.35 g Mg(NO 3 ) 2 !
1 mole Mg(NO 3 ) 2
= 0.0158moles
148.3148 g
19. A 25.5-g sample of ammonium carbonate contains _______ mol of ammonium ions.
(a). 0.468
(b). 0.288
(c). 0.531
(d). 2.00
Explanation: Realize that the formula for ammonium carbonate is (NH4)2CO3 and
calculate the molar mass (96.0856 g/mol). Convert grams to moles and then using the
stoichiometric ratio find the # of moles of ammonium ions.
1 mol (NH 4 )2 CO 3
2 moles NH +4
25.5 g (NH 4 )2 CO 3 !
!
= 0.531 moles
96.0856 g
1 mol (NH 4 )2 CO 3
20. Magnesium and nitrogen react in a combination reaction to produce magnesium
nitride:
3Mg + N2
Mg3N2
In a particular experiment, a 5.47-g sample of N2 reacts completely. How many
grams of Mg are needed for this reaction?
(a). 14.2 g
(b). 24.1 g
(c). 16.1 g
(d). 0.92 g
Explanation: Ensure that the equation is balanced. The grams of N2 must be
converted to moles of N2 and then using the stoichiometric ratio between the Mg and
N2, the grams of Mg can be calculated.
5.47 g N 2 !
1 mole N 2 3 mole Mg 24.3050 g Mg
!
!
= 14.2 g Mg
28.0134 g 1 mole N 2
1mole Mg
21. The combustion of ammonia in the presence of excess oxygen yields NO2 and H2O:
4NH3 (g) + 7O2 (g)
4NO2 (g) + 6H2O (g)
How many grams of oxygen will the combustion of 208.5 g of ammonia consume?
(a). 94.9 g
(b). 54.1 g
(c). 223.9 g
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CHE141
Chapter 3
(d).685.5 g
Explanation: Ensure that the equation is balanced and then convert the grams of
ammonia to moles. Using the stoichiometric ratio between ammonia and oxygen
calculate the grams of oxygen required. Pay attention to the molar masses being use
here.
1 mol NH 3 7 moles O 2 31.998 g O 2
208.5 g NH 3 !
!
!
= 685.5 g O 2
17.031 g
4 moles NH 3 1 mole O 2
22. The combustion of ammonia in the presence of excess oxygen yields NO2 and H2O:
4NH3 (g) + 7O2 (g)
4NO2 (g) + 6H2O (g)
How many grams of NO2 will be produced by the combustion of 208.5 g of
ammonia?
(a). 205.8 g
(b). 208.5 g
(c). 563.2 g
(d). 954.2 g
Explanation: Ensure that the equation is balanced and then convert the grams of
ammonia to moles. Using the stoichiometric ratio between ammonia and NO2
calculate the grams of NO2 required
1 mol NH 3 4 moles NO 2 46.005 g NO 2
208.5 g NH 3 !
!
!
= 563.2 g NO 2
17.031 g
4 moles NH 3
1 mole NO 2
23. Calcium carbide (CaC2) reacts with water to produce acetylene (C2H2):
CaC2 (s) + 2H2O (g)
Ca(OH)2 (s) + C2H2 (g)
How many g of H2O will be required for the production of 3.5 kg of C2H2?
(a). 4500 g
(b). 9.0 g
(c). 1800 g
(d) 4.8 x 103g
Explanation: Ensure that the equation is balanced. It is important to convert the
kilograms of C2H2 to grams first and then to moles of acetylene. Using the
stoichiometric factor between acetylene and water the grams of water can be
calculated.
1 mole 2 moles H 2 O 18.015 g
3.5 ! 10 3 g C 2 H 2 !
!
!
= 4.8 ! 10 3 g H 2 O
26.038 g 1 mole C 2 H 2
1 mole
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CHE141
Chapter 3
24. Under appropriate conditions, nitrogen and hydrogen react to yield ammonia:
N2 (g) + 3H2 (g)
2NH3 (g)
How many grams of hydrogen will be needed to react completely with 75.8 g of N2?
(a). 51.3 g
(b) 16.4 g
(c). 121 g
(d). 25.3 g
Explanation: Ensure that the equation is balanced. Convert the grams of nitrogen to
moles of nitrogen. Using the stoichiometric factor between nitrogen and hydrogen
the grams of hydrogen can be calculated.
1 mole N 2 3 moles H 2 2.016 g H 2
75.8 g N 2 !
!
!
= 16.4 g H 2
28.014 g 1 mole N 2
1 mole H 2
25. Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide:
PbCO3 (s)
PbO (s) + CO2 (g)
How many grams of lead (II) carbonate should decompose completely to produce
20.0 g of lead (II) oxide?
(a). 0.41 g
(b). 20.0 g
(c) 23.9 g
(d). 20.9 g
Explanation: Ensure that the equation is balanced. Convert the grams of lead (II)
oxide to moles of lead (II) oxide. Using the stoichiometric factor between lead (II)
oxide and lead (II) carbonate the grams of lead (II) carbonate can be calculated.
1 mole PbO 1 mole PbCO 3 267.208 g PbCO 3
20.0 g PbO !
!
!
= 23.9 g PbCO 3
223.199 g
1 mole PbO
1 mole PbCO 3
26. GeF3H is formed from GeH4 and GeF4 in the combination reaction:
GeH4 + 3GeF4
4GeF3H
If the reaction yield is 92.6%, how many moles of GeF4 are needed to produce 8.00
mol of GeF3H?
(a). 3.24
(b).6.48
(c). 5.56
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CHE141
Chapter 3
(d). 2.16
Explanation: This question is based on the ideas of actual, theoretical and percent
yields. According to the balanced equation 3 moles of GeF4 would produce 4 moles
of GeF3H if the yield was 100% (which would be the theoretical yield).
The % yield is only 92.6% meaning only 92.6% of 4 moles of GeF3H (0.926 x 4 =
3.704 moles) will be produced.
To produce 8 moles of GeF3H, theoretically 6 moles of GeF4 will be required, but
since the %yield is only 92.6%, the # of moles of GeF4 required =
8.00 moles GeF3 H ! 3 moles GeF4
= 6.48 moles GeF4
3.704 moles GeF3 H
27. When a substance burns in air, what component of air reacts?
(a). oxygen
(b). nitrogen
(c). carbon dioxide
(d). water
28. Of the reactions below, which one is not a combination reaction?
(a). C + O2
(b). 2Mg + O2
(c). CaO + H2O
(d).2CH4 + 4O2
CO2
2 MgO
Ca (OH) 2
2CO2 + 4H2O
29. Of the reactions below, which one is a decomposition reaction?
(a). NH4Cl
(b). 2Mg + O2
(c). 2N2 + 3H2
(d). Cd(NO3)2 + Na2S
NH3 + HCl
2 MgO
2NH3
CdS + 2NaNO3
30. Which of the following are combination reactions?
(1).
(2).
(3).
(4).
CH4 (g) + O2 (g)
CaO (s) + CO2 (g)
4Fe (s) + 3O2 (g)
PbCO3 (s)
CO2 (g) + H2O (l)
CaCO3 (s)
2Fe2O3 (s)
PbO (s) + CO2 (g)
(a). 1, 2, and 3 only
(b).2 and 3 only
(c). All of them
(d). 2, 3, and 4 only
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CHE141
Chapter 3
31. The mass % of Al in aluminum sulfate (Al2(SO4)3) is ________.
(a). 7.886
(b). 21.93
(c). 15.77
(d). 45.70
Explanation: Calculate the formula mass of aluminum sulfate first (= 342.143 g/mol)
and then divide the mass of the aluminum present in aluminum sulfate by the formula
mass. Don’t forget to multiply by 100%
2 ! Molar mass of Al (26.98 g/mol)
! 100% = 15.77%
Formula mass of aluminum sulfate (342.143 g/mol)
32. The formula weight of a substance is __________.
(a). the same as the percent by mass weight
(b). determined by combustion analysis
(c). the sum of atomic weights of each atom in its chemical formula
(d). the weight of a sample of the substance.
33. The mass % of C in methane (CH4) is _________.
(a). 25.13
(b). 13.36
(c). 92.26
(d).74.87
Explanation: Calculate the formula mass of methane first (= 16.042 g/mol) and then
divide the mass of the carbon present in the methane by the formula mass. Don’t
forget to multiply by 100%
' 12.01g/mol (mass of C present) $
%%
"" ! 100% = 74.87%
& 16.042g/mol (molar mass of methane) #
34. One mole of ____________ contains the largest number of atoms.
(a). S8
(b).C10H8
(c). Al2(SO4)3
(d). Na3PO4
Explanation: This question is based on the definition of a mole and the number of
atoms in a given formula. One mole of molecules = 6.022 x 1023 molecules. Since
the molecules of each substance here are made of a different number of atoms, the
one with the largest number of atoms in its formula (18 in C10H8) will be the answer.
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CHE141
Chapter 3
35. One million argon atoms is ___________ mol of argon atoms.
(a). 1.7 x 10-18
(b). 6.0 x 1023
(c). 1.7 x 10+18
(d). 1.0 x 10+6
Explanation: By definition, one mole of argon atoms = 6.022 x 1023 atoms. The
number of moles of argon atoms equal to 1 million argon atoms can then be
calculated by:
1.0 " 106 atoms of Ar
= 1.7 " 10!18 moles of Argon atoms
23
6.022 " 10 atoms/mole
36. There are __________ atoms of oxygen in 300 molecules of CH3CO2H.
(a). 600
(b). 150
(c). 3.01 x 1024
(d). 3.61 x 1026
Explanation: Since there are 2 atoms of oxygen per molecule of CH3CO2H, there are
2 x 300 = 600 atoms of oxygen present.
37. How many molecules of CH4 are in 48.2 g of this compound?
(a). 5.00 x 1024
(b). 3.00
(c). 2.90 x 1025
(d).1.81 x 1024
Explanation: Convert the grams of methane to moles of methane first and then using
the definition of a mole, calculate the # of molecules.
Since the question is asking for the number of molecules in 48.2 g of methane, the
answer will be a large number.
1 mole CH 4 6.022 ! 1023 molecules
48.2 g CH 4 !
!
= 1.81 ! 1024 molecules of CH 4
16.042 g
1 mole CH 4
38. A sample of CH2F2 with a mass of 20.0 g contains _______ atoms of F.
(a). 2.2 x 1023
(b). 40.0
(c). 4.63 x 1023
(d). 4.6 x 10-23
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CHE141
Chapter 3
Explanation: Convert the grams of CH2F2 to moles first and then using the definition
of a mole, calculate the # of molecules. Now use the number of atoms of F per mole
to find the number of fluorine atoms.
Since the question is asking for the number of atoms in 20.0 g of CH2F2, the answer
will be a large number.
1 mole CH 2 F2
6.022 ! 1023 molecules CH 2 F2
2 atoms F
20.0 g CH 2 F2 !
!
!
52.022 g CH 2 F2
1 mole CH 2 F2
1 molecule CH 2 F2
= 4.63 ! 1023 atoms F
39. How many atoms of nitrogen are in 100.0 g of NH4NO3?
(a). 3.5
(b).1.50 x 1024
(c). 1.50 x 1023
(d). 1.8
Explanation: Convert the grams of NH4NO3 to moles first and then using the
definition of a mole, calculate the # of molecules. Now use the number of atoms of N
per mole to find the number of nitrogen atoms.
Since the question is asking for the number of atoms of N in 100.0 g of NH4NO3, the
answer will be a large number.
1 mole NH 4 NO3
6.022 ! 1023 molecules of NH 4 NO3
2 atoms N
100.0 g NH 4 NO3 !
!
!
80.0424 g NH 4 NO3
1 mole NH 4 NO3
1 mole NH 4 NO3
= 1.50 ! 1024 atoms of N
40. What is the mass in grams of 9.76 x 1018 atoms of magnesium?
(a). 24.30
(b). 1.62 x 10-11
(c). 3.94 x 104
(d).None of the above
Explanation: Convert the number of atoms to number of moles and then using the
molar mass of magnesium, calculate the number of grams.
As the number of atoms is being converted to grams the answer will be a small
number.
1 mole Mg atoms 24.305 g Mg
9.76 " 1018 atoms Mg "
"
= 3.94 " 10! 4 g of Mg
23
6.022 " 10 atoms 1 mole Mg
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CHE141
Chapter 3
41. Calculate the number of sulfur dioxide molecules in 1.58 moles of sulfur dioxide.
How many molecules of oxygen is this number equal to?
(a). 1.08 x 1023
(b). 6.02 x 10-24
(c). 9.51 x 1024
(d). 1.08 x 1024
Explanation: Convert the number of moles to number of molecules of SO2 and then
to atoms of oxygen. Need to find the number of molecules of oxygen after this.
6.022 ! 1023 molecules
2 atoms of O
1 molecule O 2
1.58 moles of SO 2 !
!
!
1 mole of SO 2
1 molecule of SO 2 2 atoms of O
= 9.51 ! 1024 molecules of O 2
42. A nitrogen oxide is 30.4% by mass nitrogen. The molecular formula could be
____________.
(a). NO
(b). N2O4
(c). either NO2 or N2O4
(d). NO2
Explanation: Calculate the mass percent of nitrogen in each of the oxides based on
the molar mass of the oxide and the mass of nitrogen present in it.
14.007
For NO, mass % of N =
! 100% = 46.7%
30.006
Doing similar calculations for the rest N2O4: 30.4%, NO2: 30.4%
43. A sulfur oxide is 50.0% by mass sulfur. It’s molecular formula could be
___________.
(a). SO2
(b). SO
(c). S24
(d).Both SO2 or S2O4
Explanation: Calculate the mass percent of sulfur in each of the oxides based on the
molar mass of the oxide and the mass of sulfur present in it. SO2 is one of the correct
answers but so is S2O4, so do not stop at just the SO2.
32.065
For SO 2 , mass % of S =
! 100% = 50.0%
64.058
Doing similar calculations for the others SO = 66.7% and S2O4 = 50.0%.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 3
44. Which hydrocarbon pair below has identical mass percentage of C?
(a). C3H4 and C3H6
(b). C2H4 and C3H4
(c). C2H4 and C3H6
(d). C2H4 and C4H2
Explanation: Calculate the mass percent of carbon in each of the hydrocarbons based
on the molar mass of the hydrocarbon and the mass of carbon present in it.
24.022
For C 2 H 4 , mass % of C =
! 100% = 85.6%
28.052
Doing similar calculations for others, C3H6 = 85.6%, C3H4 = 89.9% and C4H2 =
95.9%
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