Solutions Test 1_Fall 2023
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Virginia Tech *
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Course
1035
Subject
Geology
Date
Jan 9, 2024
Type
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17
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1. What is the correct number of significant figures in the following calculation? 532.0 + 7.3 – 48.523 = a. 1 b. 2 c. 3 *d. 4 e. 5 Answer: 532.
0
+ 7.
3
– 48.
523
= 490.
7
77 The answer must have the same number of decimal places as the measurement with the fewest decimal places. Therefore, the answer should only have one digit after the decimal. The answer with one decimal place is 490.8, which has four significant figures. ______________________________________________________________________________ 2. Which of the following numbers has the most significant figures? 0.1250 4.62 × 10
6 0.00335 0.000550 512 *a. 0.1250 b. 4.62 × 10
6
c. 0.00335 d. 0.000550 e. 512 f. all have the same number of significant figures. Answer: (Signiant digits are in red) 0.
1250 4.62 × 10
6 0.00
335 0.000
550 512 *a. 0.1250 (4 significant figures). All the others have only 3 significant figures. ______________________________________________________________________________ 3. Some seafloor sediments are contaminated with high amounts of mercury. One sample of sediment contains 2.00 kg of mercury per metric ton of sediment. What mass of mercury, in kilograms, is contained in a 100. L sample of this contaminated sediment? (1 metric ton = 1000 kg; 1.70 g sediment = 1 mL of sediment) *a. 0.340 kg Hg b. 1.70 × 10
5
kg Hg c. 0.170 kg Hg d. 170. kg Hg e. 340. kg Hg Answer: *a. 0.340 kg Hg Conversion factors:
2.00 kg of mercury = 1 metric ton of sediment 1 metric ton = 1000 kg 1.70 g sediment = 1 mL of sediment kg of Hg = 0.340 kg Hg = 100.
𝐿𝐿
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚
×
1000 𝑚𝑚𝐿𝐿
1
𝐿𝐿
×
1.70 g sediment
1 mL of sediment
×
1
𝑘𝑘𝑘𝑘
1000 𝑘𝑘
×
1 metric ton
1000 𝑘𝑘𝑘𝑘
×
2.0 𝑘𝑘𝑘𝑘
𝐻𝐻𝑘𝑘
1 metric ton ______________________________________________________________________________ 4. How many liters are there 2.45 × 10
8
mm
3
? (1cm
3
= 1 mL) a. 2.45 × 10
-4
liter *b. 2.45 × 10
2
liters c. 2.45 × 10
3
liters d. 2.45 × 10
4
liters e. 2.45 × 10
5
liters Answer = *b. 2.45 × 10
2
liters = 2.45 × 10
8
𝑚𝑚𝑚𝑚
3
×
1
3
𝑚𝑚
3
1000
3
𝑚𝑚𝑚𝑚
3
×
100
3
𝑐𝑐𝑚𝑚
3
1
3
𝑚𝑚
3
×
1 𝑚𝑚𝐿𝐿
1 𝑐𝑐𝑚𝑚
3
×
1 L
1000 𝑚𝑚𝐿𝐿
= 2.45 × 10 2
liters
______________________________________________________________________________ 5. The gas particles in air are constantly in motion and have an average velocity of 464 m/s. What is the velocity of the particles in air when expressed in mi/hr? (1 in = 2.54 cm; 5280 ft = 1 mi, 12 in = 1 ft). *a. 1040 mi/hr b. 1520 mi/hr c. 464 mi/hr d. 0.288 mi/hr Answer: *a. 1040 mi/hr 464 𝑚𝑚
𝑠𝑠
×
100 𝑐𝑐𝑚𝑚
1 𝑚𝑚
×
1 𝑠𝑠𝑚𝑚
2.54 𝑐𝑐𝑚𝑚
×
1 ft
12 𝑠𝑠𝑚𝑚
×
1 mile
5280 𝑓𝑓𝑚𝑚
×
60 s
1 𝑚𝑚𝑠𝑠𝑚𝑚
×
60 min
1 ℎ𝑟𝑟
= 1040 𝑚𝑚𝑠𝑠
/
ℎ𝑟𝑟
______________________________________________________________________________
6. Carry out the following calculation and report to the correct number of significant figures. 1.54
�
23.42
−
2.3
1.248 × 10
3
�
=
*a. 0.0261 b. 0.026 c. 0.030 d. 0.02606 Answer = *a. 0.0261 Take the difference in the numerator and use the rule for addition and subtraction. The answer can have only one digit to the right of the decimal point since 2.3 has only one digit to the right. Since this is a multistep calculation, don’t round it yet, but keep track of the decimal places that should be present in this step. 1.54
�
21.1
2
1.248 × 10
3
�
=
Divide the unrounded numerator by the denominator and multiply by 1.54. Use the rule for multiplication and division to express that answer to the correct number of significant figures: 0.0261. The answer should only have three significant figures. ______________________________________________________________________________ 7. An empty flask has a mass of 251 grams. When the flask is filled with liquid ethanol, density = 0.790 g/mL, the combined mass of the flask and the liquid is 468 grams. The flask is then filled with 344 g of an unknown liquid. What is the density, in g/mL, of this unknown liquid? a. 0.621 g/mL b. 1.00 g/mL *c. 1.25 g/mL d. 2.16 g/mL Answer: *c. 1.25 g/mL Mass of the ethanol = 468 g-251 g = 217 g Volume of the ethanol = 217 𝑘𝑘
×
1 𝑚𝑚𝐿𝐿
0.790 𝑘𝑘
= 274.68 𝑚𝑚𝐿𝐿
Volume of the flask = 274.68 mL Density of the unknown liquid = 344 𝑘𝑘
274.68 𝑚𝑚𝐿𝐿
= 1.25 𝑘𝑘
/
𝑚𝑚𝐿𝐿
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______________________________________________________________________________ 8. The answer to 0.4800
2
in scientific notation is a. 0.02304 × 10
1
*b. 2.304 × 10
−1
c. 0.2304 × 10
0
d. 23.04 × 10
−2
Answer = *b. 2.304 × 10
−1
0.4800
2
= 0.4800 × 0.4800
= 0.2304 = 2.304 × 10
−1
______________________________________________________________________________ 9. A laboratory instructor gives a sample of amino-acid powder to each of four students, I, II, III, and IV, and they weigh the samples. The true value is 8.72 g. Their results for three trials are Student I: 8.72 g, 8.73 g, 8.71 g Student II: 8.56 g, 8.77 g, 8.83 g Student III: 8.50 g, 8.48 g, 8.51 g Student IV: 8.41 g, 8.72 g, 8.55 g Which student’s data is considered to have both high accuracy and high precision? *a. Student I b. Student II c. Student III d. Student IV e. None of the students have both accurate and precise data. Answer: Student I: 8.72 g, 8.73 g, 8.71 g Average = (8.72 g + 8.73 g + 8.71 g)/3 = 8.72 g. The average value is closest to the true value for Student I. Student I has the most accurate data. Student I also has measurements that are close to each other. So, Student I has the most precise data as well. ______________________________________________________________________________ 10. If the reading on thermometer A is 147.2 ºF, and the reading on thermometer B is 64. ºC, what temperature does thermometer C display if it is marked with the Kelvin scale? *a. 337 K b. – 209 K c. – 345 K d. 211 K e. 419 K
Answer: *a. 337 K Kelvin temperature = 64 o
C+ 273 = 337 K ______________________________________________________________________________ 11. Which of the following statements is not correct. a. The atomic number gives the number of electrons in a neutral atom. b. Group 17 nonmetals have charge of 1– in a binary ionic compound. c. The inner transition elements are all metals. d. Mixtures can be separated into their pure components by physical means. *e. In the compound CCl
4
, the carbon forms ions with a 4+ charge. Answer: *e. In the compound CCl
4
, the carbon forms ions with a 4+ charge. FALSE CCl
4
is a binary covalent compound. There are no ions in covalent compounds. Atoms are chemically bonded through the sharing of electrons. ______________________________________________________________________________
12. Which of the following compounds is correctly named? a. Mn
2
O
3
manganese(II) oxide b. (NH
4
)
3
PO
4
triammonium monophosphate c. FeCl
2
iron chloride d. Na
2
SO
3
sodium sulfide *e. CoN cobalt(III) nitride Answer: *e. CoN cobalt(III) nitride Review the ionic compound naming rules. b. (NH
4
)
3
PO
4
is ammonium phosphate c. FeCl
2
is iron(II) chloride d. Na
2
SO
3
is sodium sulfite *e. CoN is cobalt(III) nitride ______________________________________________________________________________ 13. If aluminum tungstate is Al
2
(WO
4
)
3
, the chemical formula for tungstic acid is: a. H
3
WO
4 *b. H
2
WO
4
c. HWO
4
d. H(WO
4
)
3
e. H
2
(WO
4
)
3 Answer: *b. H
2
WO
4
Find the charge on the tungstate ion from the known charge of aluminum: 6+ 6- 3+ 3- Al
2
(WO
4
)
3
Tungstate ion = WO
4
2
̶
Two H
+
ions are needed to balance the charge on the tungstate anion. Tungstic acid = H
2
WO
4
______________________________________________________________________________
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14. What is the formula for hydrobromic acid? a. HBrO b. HBrO
2
c. HBrO
3
*d. HBr e. HBO
4 Answer: *d. HBr Review the rules for naming acids. HBr is a binary acid. The other acids are oxoacids. ______________________________________________________________________________ 15. A certain atom has the symbol 𝑋𝑋
73
148
What does this symbol say about an atom of the element? a. It has 148 electrons. b. It is a nonmetal. c. It has 221 protons in the nucleus. *d. It has 75 neutrons in the nucleus.
e. Its atomic mass is 73 amu Answer: *d. It has 75 neutrons in the nucleus.
X has a mass number of 148 X has 73 protons. X = Ta (tantalum) Ta is a metal. X has 148-73 = 75 neutrons Atomic mass of Ta = 180.95 amu ______________________________________________________________________________ 16. Naturally occurring element X is a mixture of two isotopes: 104
X and 96
X. The isotope 104
X has a mass of 103.9 amu and an abundance of 94.6 %. The average atomic mass of X is 103.5 amu. What is the atomic mass of 96
X? *a. 96.5 amu b. 98.0 amu c. 98.3 amu d. 63.6 amu The fractional abundance = 104
X = 0.946 The fractional abundance = 96
X =1- 0.946 = 0.054
103.5 amu = 103.9 amu (0.946) + 96
X (0.054) 96
X = 96.5 amu ______________________________________________________________________________ 17. A sample of compound contains 1.65 kg of magnesium and 2.57 kg of fluorine. What mass of magnesium is in a 3.09 kg sample of the compound? a. 1.88 kg b. 1.42 kg c. 1.89 kg *d. 1.21 kg e. 2.54 kg Answer = *d. 1.21 kg Total mass of sample = 1.65 kg + 2.57 kg = 4.22 kg Mass of Mg in 3.09 kg sample = 3.09 𝑘𝑘𝑘𝑘
𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠
×
1.65 𝑘𝑘𝑘𝑘
𝑀𝑀𝑘𝑘
4.22 𝑘𝑘𝑘𝑘
𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠
= 1.21 𝑘𝑘𝑘𝑘
𝑀𝑀𝑘𝑘
______________________________________________________________________________ 18. A –3 anion has 18e
–
. The element is a. S b. Ca c. Ar d. N *e. P Answer: *e. P The atomic number (number of protons) determines the identity of the element. The –3 charge indicates the addition of 3 electrons to the neutral atom. Taking out the 3 additional electrons leaves 18 – 3 = 15 electrons in the neutral atom. This is element 15 since the # electrons = # protons = atomic number in the neutral atom. Element 15 is phosphorus
. ______________________________________________________________________________
19. The scene below represents i. a pure substance ii. a mixture iii. two elements and one compounds iv. three elements v. three compound a. i, ii, and iii are correct *b. only ii and iii are correct c. only iv is correct d. only v is correct. e. none of the statements are correct. Answer = *b. only ii and iii are correct Molecule of Element 1 Molecule of Element 2 Molecule of Compound 1
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______________________________________________________________________________ 20. In Mn
3
(PO
4
)
2
a. the Mn atom gained 2 protons. b. the Mn atom gained 3 protons. c. the Mn atom lost 3 electrons. *d. the Mn atom lost 2 electrons. e. the Mn atom has 25 electrons. Answer: the Mn atom lost 2 electrons. Find the charge on manganese using the known charge for phosphate (-3). +6 -6 +2 -3 Mn
3
(PO
4
)
2
The charge on each manganese is +2. The Mn atom loses 2 electrons to form Mn
2+
in Mn
3
(PO
4
)
2. _____________________________________________________________________________ 21. The figure below is a cupcake making kit. A student follows all directions on the box and uses up the baking kit completely. This makes 9.5 cupcakes. What was the percent yield of this baking activity? *a. 79.2% b. 20.8% c. 8.5% d. 83.3%
e. 89.3% % 𝑦𝑦𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
=
𝐴𝐴𝑐𝑐𝑚𝑚𝐴𝐴𝑠𝑠𝑠𝑠
𝑦𝑦𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑇𝑇ℎ𝑠𝑠𝑒𝑒𝑟𝑟𝑠𝑠𝑚𝑚𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠
𝑦𝑦𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
× 100
% 𝑦𝑦𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
=
9.5 𝑐𝑐𝐴𝐴𝑠𝑠𝑐𝑐𝑠𝑠𝑘𝑘𝑠𝑠𝑠𝑠
12 𝑐𝑐𝐴𝐴𝑠𝑠𝑐𝑐𝑠𝑠𝑘𝑘𝑠𝑠𝑠𝑠
× 100 = 79.2 %
_____________________________________________________________________________ 22. What mass of Fe
2
S
3
(molar mass = 208 g/mol) is produced from the reaction of 11.6 g of Fe(NO
3
)
3
(242 g/mol) with excess Na
2
S (62.0 g/mol) according to the reaction below? Fe(NO
3
)
3
(aq)
+ Na
2
S
(aq)
→
Fe
2
S
3
(s)
+ NaNO
3
(aq) (not balanced) *a. 4.99 g b. 6.75 g c. 9.97 g d. 11.6 g e. 2.33 g Answer = *a. 4.99 g Balance the reaction and use the mole ratios between Fe(NO
3
)
3 and Fe
2
S
3 to determine the mass of Fe
2
S
3. 2 Fe(NO
3
)
3
(aq)
+ 3 Na
2
S
(aq)
→
Fe
2
S
3
(s)
+ 6 NaNO
3
(aq) 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐹𝐹𝑠𝑠
2
𝑆𝑆
3
= 11.6 𝑘𝑘
𝐹𝐹𝑠𝑠
(
𝑁𝑁𝑁𝑁
3
)
3
1 ×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐹𝐹𝑠𝑠
(
𝑁𝑁𝑁𝑁
3
)
3
242 𝑘𝑘𝐹𝐹𝑠𝑠
(
𝑁𝑁𝑁𝑁
3
)
3
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐹𝐹𝑠𝑠
2
𝑆𝑆
3
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐹𝐹𝑠𝑠
(
𝑁𝑁𝑁𝑁
3
)
3
×
208 g 1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐹𝐹𝑠𝑠
2
𝑆𝑆
3
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐹𝐹𝑠𝑠
2
𝑆𝑆
3
= 4.99
𝑘𝑘
_____________________________________________________________________________
23. A compound of hydrogen, chlorine and oxygen contains 1.18% H and 42.0% Cl. What is the empirical formula of this compound? a. HCl b. HClO c. HClO
2 *d. HClO
3
e. HClO
4
Answer: *d. HClO
3 Assume 100 g of sample. Mass of H in 100 g = 1.18 g Mass of Cl in 100g = 42.0 g Mass of O in 100 g = 100 g – (1.18 g + 42.0 g) = 56.82 g 𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠𝑒𝑒𝑓𝑓
𝐻𝐻
= 1.18 𝑘𝑘
𝐻𝐻
1 ×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
1.01 𝑘𝑘
𝐻𝐻
= 1.1683 𝑚𝑚𝑒𝑒𝑠𝑠
𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠𝑒𝑒𝑓𝑓
𝐶𝐶𝑠𝑠
= 42.0 𝑘𝑘
𝐶𝐶𝑠𝑠
1 ×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶𝑠𝑠
35.45 𝑘𝑘
𝐶𝐶𝑠𝑠
= 1.1848 𝑚𝑚𝑒𝑒𝑠𝑠
𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠𝑒𝑒𝑓𝑓
𝑁𝑁
= 56.82 𝑘𝑘
𝑁𝑁
1 ×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
16.0 𝑘𝑘
𝑁𝑁
= 3.5512 𝑚𝑚𝑒𝑒𝑠𝑠
H
1.1683
Cl
1.1848
O
3.5512
Divide each mole number by the smallest mole number to obtain integers:
H
1.1683/1.1683
Cl
1.1848/1.683
O
3.5512/1.1683 Empirical formula = H
1
Cl
1
O
3 _____________________________________________________________________________
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24. The formula for methane is CH
4
(molar mass = 16.0 g/mol). This means that, in methane, a. for every 100 atoms of hydrogen there are 400 atoms of carbon. b. for every atom of carbon there are four grams of hydrogen. c. for every gram of hydrogen there are twelve grams of carbon. *d. for every one mole of CH
4
, there are 4× (6.02× 10
23
) atoms of hydrogen. Answer = *d. for every one mole of CH
4
, there are 4× (6.02× 10
23
) atoms of hydrogen. a. for every 100 atoms of hydrogen there are 400 atoms of carbon. FALSE 𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚𝑠𝑠
𝑒𝑒𝑓𝑓
𝐶𝐶
=
100 𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚𝑠𝑠
𝑒𝑒𝑓𝑓
𝐻𝐻
1 ×
1 𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚
𝐶𝐶
4 𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚𝑠𝑠
𝐻𝐻
= 25 𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚𝑠𝑠
𝐶𝐶
b. for every atom of carbon there are four grams of hydrogen. FALSE 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐻𝐻
=
1 𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚
𝑒𝑒𝑓𝑓𝐶𝐶
1 ×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
6.02 × 10
23
𝑠𝑠𝑚𝑚𝑒𝑒𝑚𝑚𝑠𝑠
𝐶𝐶
×
4 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
×
1.01 𝑘𝑘
𝐻𝐻
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
= 6.71 × 10
−24
𝑘𝑘
𝐻𝐻
c. for every gram of hydrogen there are twelve grams of carbon FALSE 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐶𝐶
=
1 𝑘𝑘
𝐻𝐻
1 ×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
1.01 𝑘𝑘
𝐻𝐻
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
4 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
×
12.01 𝑘𝑘
𝐶𝐶
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
= 2.97 𝑘𝑘
𝐶𝐶
*d. for every one mole of CH
4
, there are 4× (6.02× 10
23
) atoms of hydrogen. TRUE 4× (6.02× 10
23
) atoms of hydrogen = 4 mol H 4 mol H
= 1 mol CH
4 (mole ratio) _____________________________________________________________________________ 25. Balance the equation N
2
H
4
+ N
2
O
4
→
N
2
+ H
2
O How many moles of N
2
will be produced for every mole of N
2
O
4
that reacts? a. one b. two *c. three d. four e. five Answer: *c. three
2 N
2
H
4
+ N
2
O
4
→
3 N
2
+ 4 H
2
O _____________________________________________________________________________ 26. What is the mass of 1.08 × 10
21
molecules of aspirin (C
9
H
8
O
4
, 180.1 g/mol)? *a. 0.323 g b. 1.05 g c. 0.882 g d. 1.26 g e. 0.989 g Answer: *a. 0.323 g 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
=
1.08 × 10
21
𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑐𝑐𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
1
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
6.02 × 10
23
𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑐𝑐𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
×
180.1 𝑘𝑘
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
= 0.323 𝑘𝑘
𝐶𝐶
9
𝐻𝐻
8
𝑁𝑁
4
_____________________________________________________________________________ 27. How many moles of Al
2
O
3
can be produced from 1.00 mol of aluminum and 0.500 mol of O
2
. 4 Al(
s
) + 3 O
2
(
g
) → 2 Al
2
O
3
(
s
) a. 0.67 mol b. 0.50 mol *c. 0.33 mol d. 0.17 mol e. 0.44 mol *c. 0.33 mol This is a limiting reactant problem. Do the following to find the moles of Al
2
O
3
produced: 𝑀𝑀𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
𝑠𝑠𝑓𝑓
𝐴𝐴𝑠𝑠
𝑠𝑠𝑠𝑠
𝑚𝑚ℎ𝑠𝑠
𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚𝑘𝑘
𝑟𝑟𝑠𝑠𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚
=
1.00 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
1
×
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
4 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
= 0.500 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
𝑀𝑀𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
𝑠𝑠𝑓𝑓
𝑁𝑁
2
𝑠𝑠𝑠𝑠
𝑚𝑚ℎ𝑠𝑠
𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚𝑘𝑘
𝑟𝑟𝑠𝑠𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚
=
0.500 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
2
1
×
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
3 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
2
= 0.33 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
𝑁𝑁𝑚𝑚𝑠𝑠𝑦𝑦
0.33 𝑚𝑚𝑒𝑒𝑠𝑠
𝐴𝐴𝑠𝑠
2
𝑁𝑁
3
𝑠𝑠𝑠𝑠
𝑠𝑠𝑟𝑟𝑒𝑒𝑠𝑠𝐴𝐴𝑐𝑐𝑠𝑠𝑠𝑠
____________________________________________________________________________
28. Consider the reaction between hydrogen and oxygen. 2 H
2
(
g
) + O
2
(
g
) →
2 H
2
O(
l
) Which statement is true if 10. g of hydrogen (molar mass = 2.0 g/mol) reacts with 64 g (molar mass = 32 g/mol) of oxygen? a. The limiting reagent is hydrogen. b. 1.0 mole of oxygen will be left over after the reaction. *c. 1.0 mol of hydrogen will be left over after the reaction. d. 2.0 mol of water will form. Answer: *c. 1.0 mol of hydrogen will be left over after the reaction. This is a limiting reactant problem. Do the following to find the moles of H
2
O produced: 𝑀𝑀𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐻𝐻
2
𝑁𝑁
𝑠𝑠𝑓𝑓
𝐻𝐻
2
𝑠𝑠𝑠𝑠
𝑚𝑚ℎ𝑠𝑠
𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚𝑘𝑘
𝑟𝑟𝑠𝑠𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚
=
10.
𝑘𝑘
𝐻𝐻
2
1
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
2.0 𝑘𝑘𝐻𝐻
2
×
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
𝑁𝑁
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
= 5.0 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
𝑁𝑁
𝑀𝑀𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐻𝐻
2
𝑁𝑁
𝑠𝑠𝑓𝑓
𝑁𝑁
2
𝑠𝑠𝑠𝑠
𝑚𝑚ℎ𝑠𝑠
𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚𝑘𝑘
𝑟𝑟𝑠𝑠𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠𝑚𝑚𝑚𝑚
=
64 𝑘𝑘
𝑁𝑁
2
1
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
2
32 𝑘𝑘𝑁𝑁
2
×
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
𝑁𝑁
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
2
= 4.0 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
𝑁𝑁
𝑁𝑁𝑚𝑚𝑠𝑠𝑦𝑦
4.0 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
𝑁𝑁
𝑠𝑠𝑠𝑠
𝑠𝑠𝑟𝑟𝑒𝑒𝑠𝑠𝐴𝐴𝑐𝑐𝑠𝑠𝑠𝑠
. The limiting reactant is O
2
. Oxygen will be completely consumed during the reaction; none will be left. H
2
is in excess. To find the amount of H
2
in excess, first find the amount of H
2 that reacts with O
2
. 𝑀𝑀𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠
𝑒𝑒𝑓𝑓
𝐻𝐻
2
𝑟𝑟𝑠𝑠𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠𝑚𝑚𝑘𝑘
𝑤𝑤𝑠𝑠𝑚𝑚ℎ
𝑁𝑁
2
=
64 𝑘𝑘
𝑁𝑁
2
1
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
2
32.0 𝑘𝑘𝑁𝑁
2
×
2 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝑁𝑁
2
= 4.0 𝑚𝑚𝑒𝑒𝑠𝑠
𝐻𝐻
2
𝑟𝑟𝑠𝑠𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠
Moles of H
2
left over = 5.0 moles (available) - 4.00 moles (reacts) = 1.0 mole ____________________________________________________________________________ 29. One molecule of a hypothetical triatomic gas, G
3
, has a mass of 2.5 ×
10
–22
g. The molar mass of G
will be given by a. (2.5 ×
10
–22
) ×
(6.02 ×
10
23
) b. (2.5 ×
10
–22
) ×
(6.02 ×
10
23
) ×
(3) c. (
2
.
5
×
10
–
22
) ×
(
3
)
6
.
02
×
10
23
d. 6.02 ×
10
23
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*e. �2
.
5
×
10
–
22
�
×
�6
.
02
×
10
23
�
3
Answer: *e. �2
.
5
×
10
–
22
�
×
�6
.
02
×
10
23
�
3
Molar mass of G = 2.5 ×
10
–
22
𝑘𝑘
𝐺𝐺
3
1 𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑐𝑐𝐴𝐴𝑠𝑠𝑠𝑠
𝐺𝐺
3
×
6.02 ×
10
23
𝑚𝑚𝑒𝑒𝑠𝑠𝑠𝑠𝑐𝑐𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠
𝐺𝐺
3
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐺𝐺
3
×
1 𝑚𝑚𝑒𝑒𝑠𝑠
𝐺𝐺
3
3 𝑚𝑚𝑒𝑒𝑠𝑠
𝐺𝐺
= �
2.5 ×
10
–
22
�
×
�
6.02 ×
10
23
�
3
____________________________________________________________________________ 30. Which of the following models represent compounds having the same empirical formula? (Carbon is black, hydrogen is white, and oxygen is red). *a. B, C, and D b. A, B, and C c. D and E d. A and C
Answer: *a. B, C, and D Structure Molecular Formula Empirical Formula A C
4
H
10
O
2
C
2
H
5
O B C
2
H
4
O C
2
H
4
O C C
4
H
8
O
2
C
2
H
4
O D C
6
H
12
O
3
C
2
H
4
O E C
5
H
8
O
2
C
5
H
8
O
2
____________________________________________________________________________ 31. Answer ALL 31 questions before submitting. The Virginia Tech Honor Code prohibits cheating, plagiarism, falsification, fabrication, multiple submissions, or any other academic misconduct that leads to a student having an unfair academic advantage. It also prohibits complicity in these activities. I have read and understand the Virginia Tech Honor Code (honorsystem.vt.edu), and I promise that I will abide by the Virginia Tech Honor Code while taking this test. a. No, I do not agree. *b. Yes, I agree to abide by the Virginia Tech Honor Code.