Chapter 3

CHE141 Chapter 3

Copyright © 2006 Dr. Harshavardhan D. Bapat 1

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

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

CHE141 Chapter 3


10. When the following equation is balanced, the coefficient of nitric acid is __________.

N2O5 (g) + H2O (l) HNO3 (aq) (a). 5 (b). 2 (c). 3 (d). 4

11. The balanced equation for the decomposition of sodium azide is __________.

(a). 2NaN3 (s) 2 Na (s) + 3N2 (g) (b). 2NaN3 (s) Na2 (s) + 3N2 (g) (c). NaN3 (s) Na (s) + N2 (g) (d). None of the above

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

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 C2H2O2

58.0 g "mol#1$

2 moles O

1 mole C2H2O2

$15.99 g O

1 mole of O= 35.8 g 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:

2

2

22 CO of moles 1.183

CO g 43.99

CO mole 1CO g 52.06 =!

18. How many moles of the compound magnesium nitrate, Mg(NO3)2, are in a2.35 g

sample of this compound?

(a) 38.4 (b) 65.8 (c) 0.0158 (d) 0.0261 Explanation: This is a straight-forward conversion from grams to moles of Mg(NO3)2 which is done as follows:

CHE141 Chapter 3


s0.0158moleg 148.3148

)Mg(NO mole 1)Mg(NO g 2.35 23

23 =!

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.

( )( )

( )moles 0.531

CONH mol 1

NH moles 2

g 96.0856

CONH mol 1CONH g 25.5

324

4324

324 =!!

+

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.

Mg g 14.2Mg 1mole

Mg g 24.3050

N mole 1

Mg mole 3

g 28.0134

N mole 1N g 5.47

2

22 =!!!

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

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.

2

2

2

3

233 O g 685.5

O mole 1

O g 31.998

NH moles 4

O moles 7

g 17.031

NH mol 1NH g 208.5 =!!!

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

2

2

2

3

233 NO g 563.2

NO mole 1

NO g 46.005

NH moles 4

NO moles 4

g 17.031

NH mol 1NH g 208.5 =!!!

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.

OH g104.8mole 1

g 18.015

HC mole 1

OH moles 2

g 26.038

mole 1HC g103.5 2

3

22

222

3!=!!!!

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.

2

2

2

2

222 H g 16.4

H mole 1

H g 2.016

N mole 1

H moles 3

g 28.014

N mole 1N g 75.8 =!!!

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.

3

3

33 PbCO g 23.9PbCO mole 1

PbCO g 267.208

PbO mole 1

PbCO mole 1

g 223.199

PbO mole 1PbO g 20.0 =!!!

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

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 =

4

3

43GeF moles 6.48

HGeF moles 3.704

GeF moles 3HGeF moles 8.00=

!

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 CO2 (b). 2Mg + O2 2 MgO (c). CaO + H2O Ca (OH) 2 (d). 2CH4 + 4O2 2CO2 + 4H2O

29. Of the reactions below, which one is a decomposition reaction?

(a). NH4Cl NH3 + HCl (b). 2Mg + O2 2 MgO (c). 2N2 + 3H2 2NH3 (d). Cd(NO3)2 + Na2S CdS + 2NaNO3

30. Which of the following are combination reactions?

(1). CH4 (g) + O2 (g) CO2 (g) + H2O (l) (2). CaO (s) + CO2 (g) CaCO3 (s) (3). 4Fe (s) + 3O2 (g) 2Fe2O3 (s) (4). PbCO3 (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

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%

15.77%100%g/mol) (342.143 sulfate aluminum of mass Formula

g/mol) (26.98 Al of massMolar 2=!

!

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%

74.87%100%methane) of mass(molar l16.042g/mo

present) C of (mass 12.01g/mol=!""

#

$%%&

'

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.

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:

atomsArgon of moles 101.7atoms/mole 106.022

Ar of atoms 101.0 18

23

6!

"="

"

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.

4

24

4

23

44 CH of molecules 101.81

CH mole 1

molecules 106.022

g 16.042

CH mole 1CH g 48.2 !=

!!!

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

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.

F atoms 104.63

FCH molecule 1

F atoms 2

FCH mole 1

FCH molecules 106.022

FCH g 52.022

FCH mole 1FCH g 20.0

23

2222

22

23

22

2222

!=

!!

!!

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.

N of atoms 101.50

NONH mole 1

N atoms 2

NONH mole 1

NONH of molecules 106.022

NONH g 80.0424

NONH mole 1NONH g 100.0

24

3434

34

23

34

3434

!=

!!

!!

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.

Mg of g 103.94Mg mole 1

Mg g 24.305

atoms 106.022

atoms Mg mole 1Mg atoms 109.76 4

23

18 !"="

"""

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.

2

24

2

22

23

2

O of molecules 109.51

O of atoms 2

O molecule 1

SO of molecule 1

O of atoms 2

SO of mole 1

molecules 106.022SO of moles 1.58

!=

!!!

!

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.

46.7%100%30.006

14.007N of % mass NO,For =!=

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.

50.0%100%64.058

32.065S of % mass,SOFor 2 =!=

Doing similar calculations for the others SO = 66.7% and S2O4 = 50.0%.

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.

85.6%100%28.052

24.022C of % mass,HCFor 42 =!=

Doing similar calculations for others, C3H6 = 85.6%, C3H4 = 89.9% and C4H2 = 95.9%

Chapter 3

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