Chemistry 180 Homework Chapter 4 Name

Chemistry 180 Homework Chapter 4
•
•
Name: ____KEY________________
Write your answers in the boxed space provided.
Observe sig fig rules, and show your unrounded and rounded (boxed) answers.
1. Balance the following chemical equations by inspection.
(a)
SO 2 Cl2 + 8 HI → H 2 S + 2 H 2 O + 2 HCl + 4 I 2
(b)
FeTiO3 + 2 H 2 SO 4 + 5 H 2 O 
→ FeSO 4 ⋅ 7H 2 O + TiOSO 4
(c)
2 Fe3 O 4 + 12 HCl +3 Cl2 
→ 6 FeCl3 + 6 H 2 O + O 2
(d)
C6 H 5 CH 2 SSCH 2 C6 H 5 + 39 2 O 2 
→14 CO 2 + 2 SO 2 + 7 H 2 O
or
2 C6 H 5 CH 2 SSCH 2 C6 H 5 + 39 O 2 
→ 28 CO 2 + 4 SO 2 + 14 H 2 O
Note: The second solution for (d) is preferred!
2. Write balanced chemical equations based on the information given.
a. solid magnesium + nitrogen gas  solid magnesium nitride
b. solid potassium chlorate  solid potassium chloride + oxygen gas
c. solid sodium hydroxide + solid ammonium chloride  solid sodium chloride + gaseous ammonia
+ water vapor
d. solid sodium + liquid water  aqueous sodium hydroxide + hydrogen gas
(a)
3Mg ( s ) + N 2 ( g ) → Mg 3 N 2 ( s )
(b)
KClO3 ( s ) 
→ KCl ( s ) +
(c)
NaOH(s) + NH 4 Cl(s) 
→ NaCl(s) + NH 3 (g) + H 2 O ( g )
(d)
2 Na(s) + 2H 2 O(l) 
→ 2 NaOH(aq) + H 2 (g)
3
2
O 2 ( g ) or
2 KClO3 ( s ) 
→ 2 KCl ( s ) + 3 O 2 ( g )
3. Write balanced chemical equations to represent the complete combustion of each of the following in excess
oxygen:
a. propylene, C3H6
b. thiobenzoic acid, C6H5COSH
c. glycerol, CH2(OH)CH(OH)CH2OH
(a)
2 C3 H 6 ( g ) + 9 O 2 ( g ) → 6 CO 2 ( g ) + 6 H 2 O ( l )
(b)
C6 H 5 COSH ( s ) + 9 O 2 ( g ) → 7 CO 2 ( g ) + 3 H 2 O ( l ) + SO 2 ( g )
(c)
2 CH 2 (OH)CH(OH)CH 2 OH ( l ) + 7 O 2 ( g ) → 6 CO 2 ( g ) + 8 H 2 O ( l )
4. A commercial method of manufacturing hydrogen involves the reaction of iron and steam.
3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
a. How many grams of H2 can be produced from 42.7 g Fe and an excess of H2O(g) (steam)?
b. How many grams of H2O are consumed in the conversion of 63.5 g Fe to Fe3O4?
c. If 14.8 g H2 is produced, how many grams of Fe3O4 must also be produced?
(a)
(b)
(c)
1mol Fe 4 mol H 2 2.016 g H 2
×
×
=
2.06 g H 2
55.85g Fe 3mol Fe
1 mol H 2
1 mol Fe 4 mol H 2 O 18.02 g H 2 O
×
×
= 27.3 g H 2 O
mass H 2 O = 63.5 g Fe ×
1 mol H 2 O
3 mol Fe
55.85 g Fe
1 mol H 2 1mol Fe3O 4 231.54 g Fe3O 4
mass Fe3O 4 = 14.82 g H 2 ×
×
×
2.016 g H 2
4 mol H 2
1mol Fe3O 4
mass H 2 = 42.7 g Fe ×
= 425g Fe3O 4
5. Solid silver oxide, Ag2O, decomposes at temperatures in excess of 300 °C, yielding metallic silver and
oxygen gas. A 3.13 g sample of impure silver oxide yields 0.187 g O2(g). What is the mass percent Ag2O
in the sample? Assume that Ag2O(s) is the only source of O2(g). [Hint: Write a balanced equation for the
reaction.]
heat

→ 4 Ag ( s ) + O 2 ( g )
The following reaction occurs: 2 Ag 2 O ( s ) 
mass Ag 2 O = 0.187g O 2 ×
% Ag 2 O =
1mol O 2 2 mol Ag 2 O 231.7 g Ag 2 O
×
×
= 2.71g Ag 2 O
32.0 g O 2
1mol O 2
1mol Ag 2 O
2.71 g Ag 2 O
× 100% = 86.6% Ag 2 O
3.13 g sample
6. An excess of aluminum foil is allowed to react with 225 mL of aqueous solution of HCl (d = 1.088 g/mL)
that contains 18.0% HCl by mass. What mass of H2(g) is produced? Use this equation:
2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
2 Al ( s ) + 6 HCl ( aq ) → 2 AlCl3 ( aq ) + 3 H 2 ( g )
mass H 2 = 225 mL soln ×
3 mol H 2 2.016 g H 2
1 mol HCl
1.088 g 18.0 g HCl
×
×
×
×
= 1.22 g H 2
1 mL 100.0 g soln 36.46 g HCl 6 mol HCl 1 mol H 2
7. What are the molarities of the following solutes?
a. aspartic acid (H2C4H5NO4) if 0.405 g is dissolved in enough water to make 100.0 mL of solution
b. acetone, C3H8O, (d = 0.790 g/mL) if 35.0 mL is dissolved in enough water to make 425 mL of solution
c. diethyl ether, (C2H5)2O, if 8.8 mg is dissolved in enough water to make 3.00 L of solution
(a)
(b)
(c)
0.405g H 2 C4 H 5 NO 4 1000 mL 1mol H 2 C4 H 5 NO 4
×
×
= 0.0304 M
100.0 mL
1L
133.10 g H 2 C4 H 5 NO 4
35.0 mL C3 H 6 O 1000 mL 0.790 g C3 H 6 O
1mol
[C3H 6 O] =
×
×
×
= 1.12 M
425 mL soln
1L
1mL
58.08g C3 H 6 O
[H 2 C4 H 5 NO 4 ] =
[( C 2 H 5 )2 O] =
8.8 mg ( C 2 H 5 )2 O
3.00 L soln
×
1g
1000 mg
×
1mol ( C 2 H 5 )2 O
74.12 g ( C 2 H 5 ) 2 O
= 4.0 × 10 −5 M
8. What volume of 0.750 M AgNO3 must be diluted with water to prepare 250.0 mL of 0.425 M AgNO3?
Volume of concentrated AgNO3 solution
VAgNO3 = 250.0 mL dilute soln ×
0.425 mmol AgNO3
1 mL conc. soln.
×
= 142 mL conc. soln
1 mL dilute soln
0.750 mmol AgNO3
Note: You could have solved this problem leaving concentration in M and volume in mL, as I have
discussed in lecture. This solution demonstrates a more rigorous approach that if volume is expressed in
mL, then concentration is then expressed in mmol. If you leave the concentration in ‘M’ then the rigor is
hidden. Also, mol/L = mmol/mL. Either way gets you to the same answer!
9. Excess NaHCO3 is added to 525 mL of 0.220 M Cu(NO3)2. The substances react as follows:
Cu(NO3)2(aq) + 2 NaHCO3(s)  CuCO3(s) + 2 NaNO3(aq) + H2O(l) + CO2(g)
a. How many grams of NaHCO3(s) will be consumed?
b. How many grams of CuCO3(s) will be produced?
The balanced chemical equation provides us with the conversion factor between the two compounds.
(a) mass NaHCO3 = 525 mL solution ×
×
0.220 mol Cu ( NO3 )2 2 mol NaHCO3
1L solution
×
×
1000 mL solution
1L solution
1mol Cu ( NO3 )2
84.00694 g NaHCO3
= 19.4 g NaHCO3
1 mol NaHCO3
Note that it is necessary to convert to the volume in L here since we will perform a moles-to-moles
conversion. When we are calculating dilution, we can use the shortcut.
(b) mass CuCO = 525 mL solution ×
3
×
0.220 mol Cu ( NO3 )2
1mol CuCO3
1L solution
×
×
1000 mL solution
1L
1mol Cu ( NO3 )2
123.6 g CuCO3
= 14.3 g CuCO3
1mol CuCO3
10. A 5.00 mL sample of an aqueous solution of H3PO4 requires 49.1 mL of 0.217 M NaOH to convert all of the
H3PO4 to Na2HPO4. The other product of the reaction is water. Calculate the molarity of the Na2HPO4
solution.
The balanced chemical equation for the reaction is:
H3PO4(aq) + 2 NaOH(aq) → Na2HPO4(aq) + 2 H2O(l)
0.217 mol NaOH 1 mol HPO 24−
1
×
×
=
0.0491 L soln ×
1.06 M HPO 24− volume s/b 0.00500 L !
1 L soln
2 mol NaOH 0.005 L
11. A method of lowering the concentration of HCl(aq) is to allow the solution to react with a small quantity of
Mg. How many milligrams of Mg must be added to 250.0 mL of 1.023 M HCl to reduce the solution
concentration to exactly 1.000 M HCl?
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
We determine the amount of HCl present initially, and the amount desired.
1.023 mmol HCl
amount HCl present = 250.0 mL ×
= 255.8 mmol HCl
1 mL soln
1.000 mmol HCl
= 250.0 mmol HCl
amount HCl desired = 250.0 mL ×
1 mL soln
1mmol Mg 24.3mg Mg
mass Mg = ( 255.8 − 250.0 ) mmol HCl ×
×
= 70. mg Mg
2 mmol HCl 1mmol Mg
Again, this solution uses the rigorous approach of ‘mmol/mL.’
12. The reaction of calcium hydride and water produces calcium hydroxide and hydrogen as products. How
many moles of H2(g) will be formed in the reaction between 0.82 mol CaH2(s) and 1.54 mol H2O(l)?
The reaction of interest is: CaH2(s) + 2 H2O(l) → Ca(OH)2(s) + 2 H2(g)
Determine the moles of H2(g) that can be formed by each reactant.
2 𝑚𝑜𝑙 𝐻2
(0.82 𝑚𝑜𝑙 𝐶𝑎𝐻2 ) �
� = 1.64 𝑚𝑜𝑙 𝐻2
1 𝑚𝑜𝑙 𝐶𝑎𝐻2
2 𝑚𝑜𝑙 𝐻2
(1.54 𝑚𝑜𝑙 𝐻2 𝑂) �
� = 1.54 𝑚𝑜𝑙 𝐻2
2 𝑚𝑜𝑙 𝐻2 𝑂
The water produces less product, so H2O is the limiting reactant and 1.54 mol of H2 are formed.
13. How many grams of H2(g) are produced by the reaction of 1.84 g Al with 75.0 mL of 2.95 M HCl? [Hint:
See problem 6 for equation.]
First determine the mass of H 2 produced from each of the reactants. The smaller mass is that produced
by the limiting reactant, which is the mass that should be produced.
1mol Al 3mol H 2 2.016g H 2
mass H 2 (Al) = 1.84g Al ×
×
×
= 0.206g H 2
26.98g Al 2 mol Al 1mol H 2
2.95mol HCl 3mol H 2 2.016g H 2
1L
mass H 2 (HCl) = 75.0 mL ×
= 0.223g H 2
×
×
×
1000 mL
1L
6 mol HCl 1mol H 2
Thus, 0.206 g H 2 should be produced.
14. Titanium tetrachloride, TiCl4, is prepared by the reaction below:
3 TiO2(s) + 4 C(s) + 6 Cl2(g)  3 TiCl4(g) 2 CO2(g) + 2 CO(g)
What is the maximum mass of TiCl4 that can be obtained from 35 g TiO2, 45 g Cl2, and 11 g C?
The number of grams of TiCl4 that can be made from the reaction mixture is determined by finding the
limiting reagent, and using the limiting reagent to calculate the mass of product that can be formed.
The limiting reagent can determined by calculating the amount of product formed from each of the
reactants. Whichever reactant produces the smallest amount of product is the limiting reagent.
1 mol TiO 2 3 mol TiCl4 189.68 g TiCl4
×
×
=
83.1 g TiCl4
79.88 g TiO 2 3 mol TiO 2
1 mol TiCl4
1 mol Cl2
3 mol TiCl4 189.68 g TiCl4
45 g Cl2 ×
×
×
=
60.2 g TiCl4
70.90 g Cl2
6 mol Cl2
1 mol TiCl4
35 g TiO 2 ×
11 g C ×
1 mol Cl2 3 mol TiCl4 189.68 g TiCl4
×
×
=
130 g TiCl4
12.01 g C
4 mol C
1 mol TiCl4
Cl2 is the limiting reagent. Therefore, 6.0×101 g TiCl4 is expected.
15. Chlorine can be generated by heating together calcium hypochlorite and hydrochloric acid. Calcium
chloride and water are also formed.
a. If 50.0 g Ca(OCl)2 and 275 mL of 6.00 M HCl are allowed to react, how many grams of chlorine gas gas
will form?
b. Which reactant, Ca(OCl)2 or HCl, remains in excess, and in what mass?
The balanced chemical equation is:
Ca(OCl)2(s) + 4 HCl(aq) → CaCl2(aq) + 2 H2O(l) + 2 Cl2(g)
n Cl2 (from Ca(OCl) 2 ) = 50.0 g Ca ( OCl ) 2 ×
n Cl2 (from HCl) = 275 mL ×
1L
1000 mL
×
1mol Ca ( OCl ) 2
142.98 g Ca ( OCl ) 2
6.00 mol HCl
1L soln
×
×
2 mol Cl 2
1mol Ca ( OCl ) 2
2 mol Cl 2
4 mol HCl
= 0.699 mol Cl 2
= 0.825 mol Cl 2
70.91g Cl2
= 49.6 g Cl2
1mol Cl2
The excess reactant is the one that produces the most Cl 2 , namely HCl(aq). The quantity of excess
HCl(aq) is determined from the amount of excess Cl2(g) it theoretically could produce (if it were the
limiting reagent).
Thus, mass Cl2 expected = 0.699 mol Cl2 ×
4 mol HCl
1000 mL
V
= 0.825-0.699 ) mol Cl2 ×
×
excess HCl (
2 mol Cl 6.00 mol HCl
2
16. How many grams of commercial acetic acid (97% CH3COOH by mass) must be allowed to react with an
excess of PCl3 to produce 75 g of acetyl chloride (CH3COCl), if the reaction has a 78.2% yield?
CH3COOH + PCl3

CH3COCl + H3PO3
(not balanced)
Balanced equation: 3CH 3COOH + PCl3 → 3CH 3COCl + H 3PO3
100.0 g calculated 1mol CH 3COCl 3mol CH 3COOH
mass acid = 75g CH 3COCl ×
×
×
78.2 g produced 78.5g CH 3COCl 3mol CH 3COCl
60.1g pure CH 3COOH
100 g commercial
×
×
= 76 g commercial CH 3COOH
1mol CH 3COOH
97 g pure CH 3COOH
17. Sodium bromide, used to produce silver bromide for use in photography, can be prepared as shown (these
are consecutive reactions). How man kilograms of iron are consumed to produce 2.50 × 103 kg NaBr?
Fe + Br2  FeBr2
3 FeBr2 + Br2  Fe3Br8
Fe3Br8 + 4 Na2CO3  8 NaBr + 4 CO2 + Fe3O4
Fe + Br2
→ FeBr2
(multiply by 3)
3 FeBr2 + Br2
→ Fe3Br8
Fe3Br8 + 4 Na2CO3 → 8 NaBr + 4 CO2 + Fe3O4
3 Fe + 4 Br2 + 4 Na2CO3 → 8 NaBr + 4 CO2 + Fe3O4
Hence, 3 moles Fe(s) forms 8 mol NaBr
MassFe consumed = 2.50×103 kg NaBr ×
= 509×103 g Fe ×
1000 g NaBr
1 kg NaBr
×
1 mol NaBr
102.894 g NaBr
×
3 mol Fe
8 mol NaBr
×
55.847 g Fe
1 mol Fe
1 kg Fe
= 509 kg Fe required to produce 2.5 × 103 kg KBr
1000 g Fe
18. A mixture of Fe2O3 and FeO was analyzed and found to be 72.0% Fe by mass. What is the percentage by
mass of Fe2O3 in the mixture?
Assuming the mass of the sample is 100.0 g, the sample contains 72.0 g Fe.
g Fe 2 O3
% by mass of Fe 2 O3 =
× 100 %
g Fe 2 O3 + g FeO
Total mass of Fe in sample = mass of Fe from Fe2O3 + mass of Fe from FeO
Let m be the mass, in grams, of Fe2O3 in the mixture.
1 mol Fe 2 O3
2 mol Fe
55.85 g Fe
×
×
g Fe from Fe 2 O3 =m g Fe 2 O3 ×
159.7 g Fe 2 O3 1 mol Fe 2 O3 1 mol Fe
The mass of FeO is (100 − m) grams.
1 mol FeO
1 mol Fe 55.85 g Fe
g Fe from FeO =
(100 − m) g FeO ×
×
×
71.85 g FeO 1 mol FeO 1 mol Fe
The sum of these two expressions is equal to 72.0 g Fe. Thus:
55.85 
55.85 


72.0 =  m × 2 ×
+ (100.0 − m) ×

159.7 
71.85 


Solve for m:
m
=
73.6 g
73.6 g
% by mass of Fe 2 O3 =
× 100 % = 73.6%
100.0 g
19. Hydrogen gas, H2(g), is passed over Fe2O3(s) at 400 °C. Water vapor is formed together with a black
residue−a compound consisting of 72.3% Fe and 27.7% O. Write a balanced equation for this reaction.
We determine the empirical formula, basing our calculation on 100.0 g of the compound.
1 mol Fe
amount Fe = 72.3 g Fe ×
= 1.29 mol Fe
÷ 1.29 
→ 1.00 mol Fe
55.85 g Fe
1 mol O
= 1.73 mol O
amount O = 27.7 g O ×
÷ 1.29 
→ 1.34 mol O
16.00 g O
The empirical formula is Fe 3 O 4 and the balanced equation is as follows.
3Fe 2 O3 (s) + H 2 (g) 
→ 2 Fe3 O 4 (s) + H 2 O(g)
20. What volume of 0.0175 M CH3OH must be added to 50.0 mL of 0.0248 M CH3OH so that the resulting
solution has a molarity of exactly 0.0200 M? Assume that the volumes are additive.
Let V be the volume of 0.0175 M CH3OH(aq) that is required.
moles CH3OH in solution C
(V + 0.0500) × 0.0200 M
= moles CH3OH in solution A + moles CH3OH in solution B
=
(V × 0.0175 M)
+
(0.050 × 0.0248 M)
Solve for V: V = 0.0960 L = 96.0 mL
You could also have solved this by letting V = volume of the final solution (in L), and then V−0.0500
would be the volume of the 0.0175 M solution needed.

Download Report