Download Exam 2 Key - Ars

Chemistry 68 - Exam 2
14 July 2016
Name _________________________
Show all work for credit. State any assumptions made to solve a problem.
Give all numerical answers with the correct number of significant figures and
the correct units. All answers in scientific notation must be in correct
scientific notation (i.e., 6.0221023 not 6.022E23 or 6.022e23). Each instance
of incorrect scientific notation will result in the loss of 3 points. All numbers
that require units should have the units written. All instances of numbers
without units will result in the loss of 3 points each.
1. (2 points each) Balance the following chemical equations:
a. 2 Na2O2 + 2 H2O  4 NaOH + O2
b. 2 NaOH + Cl2  NaCl + NaClO + H2O
c. 4 Si2H3 + 11 O2  8 SiO2 + 6 H2O
d. H2SO4 + 8 HI  H2S + 4 I2 + 4 H2O
2. (28 points) A binary compound of magnesium and nitrogen is analyzed, and 1.2791 g of the
compound is found to contain 0.9240 g of magnesium. When a second sample of this
compound is treated with water and heated, the nitrogen is driven off as ammonia, leaving
a compound that contains 60.31% magnesium and 39.69% oxygen by mass. Calculate the
empirical formulas of the two magnesium compounds. Write the chemical equation that
occurs between the first magnesium compound and water. Calculate the mass of ammonia
produced in the reaction.
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀
= 0.038017 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀
24.3050 𝑔𝑔 𝑀𝑀𝑀𝑀
1 𝑚𝑚𝑜𝑜𝑙𝑙 𝑁𝑁
(1.2791 𝑔𝑔 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 0.9240 𝑔𝑔 𝑀𝑀𝑀𝑀) ×
= 0.025352 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁
14.0067 𝑔𝑔 𝑁𝑁
0.025352 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁
0.038017 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀
= 1 𝑁𝑁 × 2 = 2
= 1.49955 ≅ 1.5 𝑀𝑀𝑀𝑀 × 2 = 3
0.9240 𝑔𝑔 𝑀𝑀𝑀𝑀 ×
0.025352 𝑚𝑚𝑚𝑚𝑚𝑚
0.025352 𝑚𝑚𝑚𝑚𝑚𝑚
The first compound is Mg3N2.
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀
= 2.48138 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀
24.3050 𝑔𝑔 𝑀𝑀𝑀𝑀
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂
39.69 𝑔𝑔 𝑂𝑂 ×
= 2.48072 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂
15.9994 𝑔𝑔 𝑂𝑂
2.48138 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀
2.48072 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂
= 1.00027 ≅ 1
=1
60.31 𝑔𝑔 𝑀𝑀𝑀𝑀 ×
2.48072 𝑚𝑚𝑚𝑚𝑚𝑚
2.48072 𝑚𝑚𝑚𝑚𝑚𝑚
The second compound is MgO. The chemical equation is:
Mg3N2 + 3 H2O  2 NH3 + 3 MgO
The amount of ammonia produced is:
1.2791 𝑔𝑔 𝑀𝑀𝑀𝑀3 𝑁𝑁2 ×
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀3 𝑁𝑁2
2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3
17.0305 𝑔𝑔 𝑁𝑁𝑁𝑁3
×
×
= 0.4137 𝑔𝑔 𝑁𝑁𝑁𝑁3
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3
100.9284 𝑔𝑔 𝑀𝑀𝑀𝑀3 𝑁𝑁2 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀3 𝑁𝑁2
3. (20 points) Calculate the following. SHOW ALL WORK FOR FULL CREDIT. (Conversion factors
are on the last page.)
a. The number of fluorine atoms in 35.00 g of C3H4F2.
35.00 g C3H4 F2 ×
1 mol C3H4 F2
2 mol F
6.022141 × 10 23 at F
5.400 10 23 at F
×
×
=×
78.0627 g C3H4 F2 1 mol C3H4 F2
1 mol F
b. Calculate the number of grams of iron in a 250.0 mL of steel that has a density of
7.84 g mL−1 and is 95.5% Iron by mass.
250.0 mL steel × ×
7.84 g steel
95.5 g Fe
×
= 1.9 × 103 g Fe
1 mL steel 100.0 g steel
c. Calculate the volume, in in3, of silver containing 6.1335×1022 atoms of silver. The
density of silver is 10.3 g/cm3.
6.1335 × 10 22 at Ag ×
1 mol Ag
107.8682 g Ag 1 cm3 Ag
×
×
1 mol Ag
10.3 g Ag
6.022141 × 10 23 at Ag
3
 1 in

×
=
6.48 × 10 −2 in3

 2.540 cm 
d. The Juno spacecraft which just entered orbit around Jupiter this past 4th of July
approached Jupiter at a speed of 70.0 km s-1. What is this in furlongs per week?
70.0 km 60 s 60 min 24 h 7 d 103 m 1 cm
1 in
1 ft
1 mi
8 fl
×
×
×
×
×
× −2
×
×
×
×
s
1 min
1h
1 d 1 wk 1 km 10 m 2.540 cm 12 in 5280 ft 1 mi
8
= 2.10 × 10 fl/wk
4. (19 points) 163.883 g of manganese(VI) chloride reacts with ammonium oxalate in a double
replacement reaction. Calculate the number of grams of solid produced if there is a 96.10
% yield.
MnCl6 + 3 (NH4)2C2O4  6 NH4Cl + Mn(C2O4)3
163.883 g MnCl6 ×
1 mol Mn ( C2 O4 )3 583.0520 g Mn ( C2 O4 )3
1 mol MnCl6
×
×
267.656 g MnCl6
1 mol MnCl6
1 mol Mn ( C2 O4 )3
×
= 343.074 g Mn ( C2 O4 )3
96.10 g Mn ( C2 O4 )3 actual
100.00 g Mn ( C2 O4 )3 theoretical
5. (25 points) Calculate the amount of heat transferred (including the sign) when 40.00 g of
1-decanol (C10H21OH) reacts with 40.00 g of oxygen gas in a combustion reaction. It is an
exothermic reaction and the absolute value of ΔH is 6601 kJ mol−1
C10H21OH + 15 O2  10 CO2 + 11 H2O ΔH = –6601 kJ
40.00 g C10 H21OH ×
40.00 g O2 ×
1 mol C10 H21OH
1 mol rxn
−6601 kJ
×
×
=
−1668 kJ
158.2811 g C10 H21OH 1 mol C10 H21OH 1 mol rxn
1 mol O2
1 mol rxn −6601 kJ
×
×
=
−550.1 kJ
31.9988 g O2 15 mol O2 1 mol rxn
–550.1 kJ of heat will be transferred.
Conversion Factors
1 qt = 0.9463 L
1 lb. = 453.6 g
1 in = 2.540 cm
2 pt = 1 qt
1 gal = 4 qt
2 c = 1 pt
8 fl. oz. = 1 c
8 furlongs = 1 mi
12 in = 1 ft
1 mi = 5280 feet
8 drams = 1 fl. oz.
1 stone = 14 lb