Chapter 3 Suggested Homework Answers

 

Problems 10, 12, 14, 16, 20, 22, 26, 34, 36, 38, 40, 44, 46, 48, 50, 58, 60, 62, 64, 72, 76, 78, 80

 

10.  (a)  If you add a subscripted “2” to CO, this makes the formula into carbon dioxide, CO2, a different compound.  If you add a coefficient “2” in front of CO, this refers to two carbon monoxide molecules, 2 CO.

(b)  In order to consistent with the law of conservation of mass, the numbers of each type of atoms must be equal on both sides of the reaction.  The reaction is:

3 Mg(OH)2(s)  +  2 H3PO4(aq)  ¾¾®  Mg3(PO4)2(s)  +  6 H2O(l)

In this reaction, there are 3 Mg atoms on the left side and 3 Mg atoms on the right side.  There are 12 H atoms on the left side (=3´2 (from 3 Mg(OH)2) + 2´3 (from 2 H3PO4)) and 12 H atoms on the right side (= 6´2 from 6 H2O).  There are 14 O atoms on the left side (= 3´2 (from 3 Mg(OH)2) + 2´4 (from 2 H3PO4)) and 14 O atoms on the right side (=2´4 (from Mg3(PO4)2) + 6 (from 6 H2O)).  Finally, there are 2 P atoms on the left side and 2 P atoms on the right side.

 

12.  Balance the following equation:

(a)  Li(s)  +  N2(g)  ¾¾®  Li3N(s)

Assuming there is a 1 in front of Li3N, this would place 1 in front of Li and ½ in front of N2:

Li(s)  +  ½ N2(g)   ¾¾®  Li3N(s)

Multiply the entire equation by 2 to get rid of the fraction:

2 Li(s)  +  N2(g)  ¾¾®  2 Li3N(s)

(b)  La2O3(s)  +  H2O(l)  ¾¾®  La(OH)3(s)

Assuming there is a 1 in front of La2O3, this would place 2 in front of La(OH)3:

 La2O3(s)  +  H2O(l)  ¾¾®  2 La(OH)3(s)

This would produce 6 H’s on the right side.  To produce 6 H’s on the left side, place a 3 in front of H2O:

La2O3(s)  +  3 H2O(l)  ¾¾®  2 La(OH)3(s)

This equation is balanced.  There are 6 O atoms on both sides.

(c)  NH4NO3(s)  ¾¾®  N2(g)  +  O2(g)  +  H2O(g)

Assuming there is a 1 in front of NH4NO3, this would place 1 in front of N2, 2 in front of H2O, and ½ in front of O2:

NH4NO3(s)  ¾¾®  N2(g)  +  ½ O2(g)  + 2 H2O(g)

Multiply the equation by 2 to get rid of the fraction:

2 NH4NO3(s)  ¾¾®  2 N2(g)  +  O2(g)  + 4 H2O(g)

(d)  Ca3P2(s)  +  H2O(l)  ¾¾®  Ca(OH)2(aq)  +  PH3(g)

Assuming there is a 1 in fron of Ca3P2, this would place 3 in front of Ca(OH)2 and 2 in front of PH3:

Ca3P2(s)  +  H2O(l)  ¾¾®  3 Ca(OH)2(aq)  +  2 PH3(g)

Now there are 12 H atoms on the right side.  To get 12 H atoms on the left side, place 6 in front of H2O:

Ca3P2(s)  +  6 H2O(l)  ¾¾®  3 Ca(OH)2(aq)  +  2 PH3(g)

Check the number of O atoms on both sides.  There are 6 O atoms on both sides, this equation is balanced.

(e)  Ca(OH)2(aq)  +  H3PO4(aq)  ¾¾®  Ca3(PO4)2(s)  +  H2O(l)

Assuming there is a 1 in front of Ca3(PO4)2, this would place 3 in front of Ca(OH)2 and 2 in front of H3PO4:

3 Ca(OH)2(aq)  +  2 H3PO4(aq)  ¾¾®  Ca3(PO4)2(s)  +  H2O(l)

Now there are 12 H atoms on the left side.  To obtain 12 H atoms on the right side, place a 6 in front of H2O:

3 Ca(OH)2(aq)  +  2 H3PO4(aq)  ¾¾®  Ca3(PO4)2(s)  + 6 H2O(l)

Check the number of O atoms on both sides to see if the equation is balanced.  There are 14 O atoms on both sides, this equation is balanced.

(f)  AgNO3(aq)  +  Na2SO4(aq)  ¾¾® Ag2SO4(s)  +  NaNO3(aq)

Assume there is a 1 in front of Ag2SO4.  This would place 2 in front of AgNO3 and 1 in front of Na2SO4:

 2 AgNO3(aq)  +  Na2SO4(aq)  ¾¾® Ag2SO4(s)  +  NaNO3(aq)

The equation is balanced by placing 2 in front of NaNO3:

2 AgNO3(aq)  +  Na2SO4(aq)  ¾¾® Ag2SO4(s)  +  2 NaNO3(aq)

(g)  CH3NH2(g)  +  O2(g)  ¾¾®  CO2(g)  +  H2O(g)  +  N2(g)

Assume there is a 1 in front of CH3NH2, this would place 1 in front of CO2, 5/2 in front of H2O and ½ in front of N2.  Multiply all these coefficients by 2 to get whole numbers:

2 CH3NH2(g)  +  O2(g)  ¾¾®  2 CO2(g)  +  5 H2O(g)  +  N2(g)

There are a total of 9 O atoms on the right side.  Place a 9/2 in front of O2 to balance the number of O atoms on both sides:

2 CH3NH2(g)  + 9/2 O2(g)  ¾¾®  2 CO2(g)  +  5 H2O(g)  +  N2(g)

Multiply the entire equation by 2 to get rid of the fraction:

4 CH3NH2(g)  + 9 O2(g)  ¾¾®  4 CO2(g)  +  10 H2O(g)  +  2 N2(g)

 

14.  (a)  Sulfur trioxide gas is SO3(g) .  Water is H2O(l).  Sulfuric acid is H2SO4(aq).   The unbalanced reaction is:

SO3(g)  +  H2O(l)  ¾®  H2SO4(aq)

This equation is already balanced.

(b)  The unbalanced reaction is:

B2S3(s)  +  H2O(l)  ¾®  H3BO3(aq)  +  H2S(g)

Assuming a “1” in front of B2S3, this would result in coefficients of 2 in front of H3BO3, 3 in front of H2S and 6 in front of H2O:

B2S3(s)  +  6 H2O(l)  ¾®  2 H3BO3(aq)  +  3 H2S(g)

(c)  The unbalance reaction is:

PH3(g)  +  O2(g)  ¾¾®  H2O(l)  +  P4O10(s)

Assume there is a “1” in front of P4O10.  This would result in a 4 in front of PH3.  This would place a 6 in front of H2O.  Complete balancing the reaction by placing a 8 in front of O2:

4 PH3(g)  +  8 O2(g)  ¾¾®  6 H2O(l)  +  P4O10(s)

(d)  The unbalanced reaction is:

Hg(NO3)2(s)  ¾¾®  HgO(s) + NO2(g)  +  O2(g)

Assume there is a “1” in front of Hg(NO3)2.  This would place a 2 in front of NO2 and a ½ in front of O2:

Hg(NO3)2(s)  ¾¾®  HgO(s) + 2 NO2(g)  + ½  O2(g)

Multiply the entire equation by 2 to remove the fraction:

2 Hg(NO3)2(s)  ¾¾®  2 HgO(s) + 4 NO2(g)  + O2(g)

(e)  The unbalanced reaction is:

Cu(s)  +  H2SO4(aq)  ¾¾®  CuSO4(aq)  +  SO2(g)  +  H2O(l)

Place a 2 in front of H2SO­4 and in front of H2O to balance this reaction:

Cu(s)  +  2 H2SO4(aq)  ¾¾®  CuSO4(aq)  +  SO2(g)  +  2 H2O(l)

 

16.  (a)  Calcium forms ions of +2 charge because it is in group 2A and oxygen forms ions of –2 charge because it is in group 6A.  The formula of the salt is CaO.  The unbalanced reaction is:

Ca(s)  +  O2(g)  ¾¾®  CaO(s)

The balanced reaction is:

2 Ca(s)  +  O2(g)  ¾¾®  2 CaO(s)

(b)  When a compound containing C, H, and O is completely combusted it forms CO2(g) and H2O(g).  The unbalanced reaction for the combustion for C3H6O(l) is:

C3H6O(l)  +  O2(g)  ¾¾®  CO2(g)  +  H2O(g)

Assuming there is a “1” in front of C3H6O, this will place a 3 in front of CO2, 3 in front of H2O, and 4 in front of O2:

C3H6O(l)  +  4 O2(g)  ¾¾®  3 CO2(g)  +  4 H2O(g)

 

20.  (a)  C3H6(g)  +  O2(g)  ¾¾®  CO2(g)  +  H2O(g)

Assume there is a “1” in front of C3H6.  This will place a 3 in front of CO2, a 3 in front of H2O, and a 9/2 in front of O2:

C3H6(g)  +  9/2 O2(g)  ¾¾®  3 CO2(g)  +  3 H2O(g)

Multiply the whole equation by 2 to get rid of the fraction:

2 C3H6(g)  +  9 O2(g)  ¾¾®  6 CO2(g)  +  6 H2O(g)

This is a combustion reaction.

(b)  NH4NO3(s)  ¾¾®  N2O(g)  +  H2O(g)

To balance this reaction place a “2” in front of H2O:

NH4NO3(s)  ¾¾®  N2O(g)  +  2 H2O(g)

This is a decomposition reaction.

(c)  C5H6O(l)  +  O2(g)  ¾¾®  CO2(g)  +  H2O(g)

Assume there is a “1” in front of C5H6O.  This will place a 5 in front of CO2, a 3 in front of H2O, and a 6 in front of O2:

C5H6O(l)  + 6 O2(g)  ¾¾®  5 CO2(g)  +  3 H2O(g)

This is a combustion reaction.

(d)  N2(g)  +  H2(g)  ¾¾®  NH3(g)

Place a 3 in front of H2 and a 2 in front of NH3 to balance this reaction:

N2(g)  +  3 H2(g)  ¾¾®  2 NH3(g)

This is a combination reaction.

(e)  K2O(s)  +  H2O(l)  ¾¾®  KOH(aq)

Place a 2 in front of KOH to balance this reaction:

K2O(s)  +  H2O(l)  ¾¾® 2 KOH(aq)

This is a combination reaction.

 

22.  (a)  Formula weight of N2O:

The formula weight is 44.02 amu.

(b)  Formula weight of HC7H5O2:

The formula weight is 110.02 amu.

(c)  Formula weight of Mg(OH)2:

The formula weight is 58.33 amu.

(d)  (NH2)2CO

The formula weight is 102.14 amu.

(e)  CH3CO2C5H11

The formula weight is 130.18 amu.

 

26.  (a)  The mass % of C in CO2:

(b)  The mass % of C in CH3OH:

(c)  The mass % of C in  C2H6:

(d)  The mass % of C in (NH2)2CS:

 

34.  (a)  Calculate the mass, in grams, of 1.906 × 10-2 mol of BaI2.

Multiply the number of moles by the molar mass of BaI2:

(b)  Calculate the number of moles in 48.3 g of NH4Cl.

Divide the mass in grams by the molar mass of NH4Cl:

(c)  Calculate the number of molecules in 0.05752 mol HCHO2.

Multiply the number of moles by Avogadro’s number:

(d)  Calculate the number of O atoms in 4.88 × 10-3 mol Al(NO3)3.

First, calculate the number of formula units by multiplying the number of moles by Avogadro’s number:

Since there are 9 O atoms in one formula unit, multiply this by 9:

 

42.  First, convert the mg of THC to g:

Divide this by the molar mass of THC (C21H30O2) to obtain the number of moles:

 

44.  (a)  Since you already have the number of moles of each element in the compound, divide each number by the smallest number in the list to obtain the smallest whole number ratio between the numbers:

The empirical formula is K2CO3.

(b)  Given the mass of each element, first convert each mass into number of moles then divide by the smallest number in the list to obtain the smallest whole number ratio:

The empirical formula is SnF4.

(c)  Assume 100 grams of compound.  Calculate the mass of each element in 100 grams of compound from the mass percents.  Divide each mass by the atomic weight of the element to calculate the number of moles of each element in 100 grams of compound.  Divide each number of moles by the smallest number of moles to determine the smallest whole number ratio:

The empirical formula is Na3AlF6.

 

46.  (a)  Assuming 100 grams of compound:

The empirical formula is K3PO4.

(b)  Assuming 100 grams of compound:

The empirical formula is Na2SiF6

(c)  Assuming 100 grams of compound:

The empirical formula is C12H12N2O3.