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Unformatted text preview: bunIce 0131323713) ~ Homework 5  3mm.  {53129) 3 This printvwl. should have 18 questions.
Multiplochoice quwions may continue on
Lllc next column or page # ﬁnd all choices before answering. 001 10.0 paints:
For the reaction 2C0+ ng2002, whle is the mammal amount of CD; which
could be formed from 13.27 H of CO and
0.1?8 mol of 0:? Correct. answer: 15.6674 .3. Explanation:
mac, m 13.27 1: no, m: 0.178 mol
This is a limiting Ieactant. problem became
the ammmts of mom than am: reactant are
gime Either CO or 02 will limit, the amount.
of (102 that can form.
Assume that CO is aha limiting rmnlmt.
The mount. of C02 that. can be produced is , 1 mol CO
1’ L100: M1327ch X 21ml C02 x 2.3.0100
K 44.0095 .1: co. 1 mol C02
2 20.34985; 00:. Assume that 02 is the reactant.
The amount of C02 that can be produced ix E‘gCOa $0.3?3m0102 x m
ImolOz x 44.90% g C103
1 ma} 002
= 35.65743 (10;. . Since a smaller arnmmt of C02 can be pro
duced with the given amount of 02, 0;. is the
limiting reagent, and a maximum $136674 g
of 00:. canbe produced. 092 10.0 points If the reaction of 1.00 mole Niigﬁg) and 1.00
mole (Mg) WW) + 502(2) —» woos) + 0 511.0(3) ‘ is eonled ow. to compleﬁon,
1. .11] of the NH; i3 consmned. ‘2. .1110? Lin: NH: is «mamnod and 1.5 mol of
1120 is produced. 3. ail afﬁne 02 is consumed. correct II. all of the N33 is unmarried and 4.0 mol of
NO(g) 'm pmtiuced. 5. 1.5 mol of HaOis produced. 6. all ofthe 02 is temmed and 1.5 moi of
N901) is produced. 7. 4.011101 nfNCXg) is produced and 1.5 fuel
of H20 is proﬁuwd. 8. all. ofthe 02 is named and 3.5 mol of
H20 is produced. 9. None of the oLluzr answms ismrrect. .10. All nftlm O; hummuan and 4.01110] of
Nag} is pruduemi. Explanation:
mm, m 1.01310] no, a 1.0 mol We recognize. this as a reactant
problem because the amounts of mm ﬁlm
one reactant are gin.11. We must. (Idamine
which of these WOLﬂd be used up first (the
limiting wagLani). To do this we compare
the mind with of reactants to the available.
mix}; of wackmts. The balanced dismal equation shows that.
we. meal 4 mal NH; fnr every limo] ()9. “’9
use that: cadﬁdentsto calculate the mqm‘mi
miioal' reactants: “ml NH; A 08111013133 5 3210102 1 mol 0; Elem mil; ratio wesee that. (men mole of ()2
that mudswquirm exactly 05 mo! Nils. Next. burke (jprE'lﬁ) — Homework 5 A4 Sutcliﬁ'e "— (53128) 3 Correct answer: 40.?4” kPa. Explanation:
“3139(an VamB‘Ich
P; 25.55 kP‘a 17,. a. '5
Applying Boyle’s law.
nmmaw
P2 M : (25.51.13.963!) cmi)
V2 87 m3 =2 40.?414 lcPa ($06 10.0 points
Which of the following is a true Wt
about a ﬁxed amount of ideal gas at. constant
preemre? 1. If a sample cigar. is (:00le from 400°C to
200°C, Lhe volume will dmem by a factor of
two. 2. If a sample of gas is heated from 100°C Lo
200°C, Um volume will decrease by a faster or
LWO. 3. If a. sample of gas is heated {mm 0°C to
273°C, the volume will double. (towed. Explanation:
Volumeis directly prupodional to temper»
2mm: exprmed on the Kelvin scale.
0°C ~+ 270"C:
0°C + 273 x 2'53 K;
273“C + 273 516 K.
The Kelvin Lamp matuxe chang by afacwr of 2 (L3,, 374:), so the volume would also double.
400°C —> 200°C:
400°C ~4 273 =2 673 K;
200°C Kb 273 m 477:3 K.
The Kelvin tmpm‘ahzxc changes by a fan mr of 0.70 (fa, , so time volume wnuld dzangcby an actor of 0.70. A temperature mch from 100°C to
200°C would cause the voiume to increase,
not decrease, use this rel:po 111le be incor HEEL 007 3.0.0 poian
At. ST? 1 gas occupics 121 mL. How may will this gas occupy at — 2°C and
1&8 arm? I. 33.0mL curlact
2. 115 mL
3. 12? mL 4. »~i9.5 ml. Explainglij
P; z 1 atm T] w 273.15 K
T9, 22 ~52?) + 273.35 3: 22115 K an_&m T: A T2
_HWB
V“ ma m (1 am) (121 mi.) (22115 K)
' (273.15 105.38 31.111)
m 83.0212 ml. (am; 10.0 points
What. mlumciﬁ occupiai by 1.00 kg of helium
at ROTC :n. a presume (£735 Torr? 1. 5.6{)x 103 L
2. 5.9m 10‘S L
3. 2.95 5410“l L
4. 1.06 x m” L 5. 5.90 x 103 L (mier Explanalion: l H 1090 g 1 mo e e
0’00!“ He) 1 kg 4.0025 :5 He = 249.338 mo] He
T : 530°C + 273.15 : 278.15 K 1 mm
Pm (was for.) mfg; # U.§é‘T.i65 aim bankn. (jpb2376) M Homework 5 — S‘atcliﬁ‘e — (53120} 2 we calculate the available mm 6f mama:
from our data: "lino! NH:
1 mo! 03 We have 1 mol of NH3 available for each
1110100! 02, mwehzvemore MzthHg
£90 II‘XH‘J. all of the 0:. We will ram out of 02
ﬁrst, so 0:; hmeﬁmﬂmgmctam. Therefore
itishuwhhatallcftheﬂg Willbccommmed
in ﬁlm Icaﬂﬁoa. We can calculate the amounts of N0 and
H70 that would be produced to check that,
the other answers are incon’ecs. Since the
rmdjtm must Mop once we run out of 02, we
me the ammut. of 02 as the basis for the5c
summons. “’2 use the mole ratios from the
chemical equation to calculxm moics N0 and
H20 produced: 4 Incl N0
5 1110102 6 mol H20
5 rrml 02 ?mol N0 :1: 1.0 531030; x a 0.80 mol N0
'5' mo] H20 9: 1.0 mol 02 X
a 1.2 mol H20 003 20.6 point.2
Consider the reaction CRCN2 “I” 3H30 m! CaCO: 4 2Nl‘l3 .
This reaction has a 75.6% yielﬁ. How many
moles of CaCNg are needed to obtain 18.5 g
of min? 1. 478 mol
2. 1.65 mo}
3. 0.4M mol
4. 0.547 ma}
5. 209 11101 6. 0.724 mm] urnact % yicid a 75.6% mm 2 18,5 3
18.6 5; NH: is; the actual yield. This is
divided by the percent. yielc? to give the theo
retical yield.
EGO ’1 a,— ..........
. moi CaCNg 13.63 NEE; x 75’s R imoiNHg
173 N55;
Ema! CaCN2 2 mol N33
= 0.724 mail 004 10.0 points
A.an contains F9504 and no other iron. The
iron in a 36.5 gram sample of the ore is :le
converted by .1 swim of chemical reactions to FegOs. The mam of FegOg is measured to be
25 mama. “mat. wag the percent. R304 in the sample ofore? Comm answar: 66.205. Explanation: mm a 35.5 g
meoa = 159.658 g/mul FWI‘oao. 2 231.533 mime]
The reactionis probably iilce Lhis: 4 FeaOa + 02 + 6 FRQOJ The key is the 2:8 ratio of the 2 iron oxldm.
Calculate the moan}. of F2304 from which
the Fmog me from: 1 mo] 2 Incl 231.5339; 159.688 .5 3 11101 X 1 mol 24.1651 of F9304 in the sample 24.1653 9
36.5 x 100/“ 66.265736 of 9930.; Who. 3 25 g 25gb: pancent. cm in the sample. 005 10.0 points
A ﬂask contaixﬁng 139 m3 of hydrogen was
collected under apxeamte of 25.5 kPa. What.
presmm would have been regulated {or the vol.
ume ofthe gas to have been 87 (:ms. amxning Explanation: the same temperature?
burke (ipb2375) w Homework 5 M Sutclin‘e v (53120} a
4. 1.35 g  L"
PV 2 nRT
V w “RT 5. 3.39 g  L‘l coxrad;
p
B 1 ti 
_ (249.838m01)(0.03205 Egg?) :9 am on 1 .1141:
w 0.952.105 mm m. (850 Tm) {mo n m 1.11842 atm
X {273.15 K) T 3 12C 4* 273.15 x 235.15 K
= 58965 L MMcl 70.906 g/mol 009 10.0 points
ComEdmEng the ideal gas law PV 2 nRT,
what is P directly proportional no? Lonly V 2. T and V
3.0aly'1" 4. n and 'I' correct 5. only 7: Explazmtiun:
Mange the aquatic): so Limb P is: the
subg'ect of the fernaxis: nRT Pam R is the gas consMm and dm not appear in
any oftherespom V is in the denominator;
1’ is invm‘sety proportional to V. 'I‘ and n are
in 1212:: numerator; P is directly proportional
when: 1: and T, mammasLbcnumber 01"
moles of a gas or the temperamre of a gas
inucases, the pressure will also mm by
the same factor. om mo [302313
What. is the (Equity of chlorine gas at 850 Tan
and 12° C?
1. 7.405; L—E
2. 27.3 317: 3.3.70g L“‘ The ideal gas law is PvanT
33' P
V LET with unit of measure moi/l. on each side.
Mul‘iplying cankby molar mam (MM) gives Tl. P
T’  MM m if  MM,
which now 1333 units uf g/L (z demity). Thus
P
W WW
1.11342 nun (0.05206 L‘ atm/mol/K) (235.15 K)
x (70.906 g/mol)
2 3.33909 3,“. c: 011 10.01)an
The empirical formula of a gas is CliaO. If
2.77 g of the gas; occupies 1.00 L at achLly
CFC at a meme at 760 ‘i‘orr. what is the
molecular formula of the gas? I . 61130 2. Cal[501 coxred.
3. C4H1204 4. CzHoOn 5. came. Explanation:
m = 2.71 g V w 1 L
'1" m 0°C w 273.15 K 1 atm P I: (765 Torr} m L" 1 aim burke (b52376) » Hamework 5 ~ 51mm m {53120} 5 The density of the sample is Pregame dungeis dunemole change. is similar (£0 Chatlcs’ Law a:ch that. pres pﬁ E M. a. 25;? glrL Blue is directly propozﬁunzl tumult»; (keeping V i E. volumc and tempermuxc consult). So that... The ideal gas inw is PanRT
11. P '17 m E?
with mail. of measure moijL on eatJ: aide.
Multiplying each by molar mass (MM) give; n P
V.MM MM 2?,
with units of glL.
MM m P————’;T m 3.11 “5133038206 L aLm/mol/KZ W 1 atm
x (273.15 K) m mama/mo: This the ACTUAL MM of the gas. An
empirical formula of N02 giwm a mum of
31.039 g/mol. DiVidlnp; this into the actual
MM ﬁve; 2.00067. 50 the ACTUAL molecu
lar farmlsin is twice the EMPIRICAL fm'mula: CgHsog. 012 10.0 poian
An oxygen tank i:ch aL 200°C contain: 28.0
moles of oxygen and the gauge reads 31.0
atm. AliAr two weeks, the gauge reads 20.1
Atm. ﬂow much mam was used during the mowed: period?
1. 15.7 mol
2. 20.5 mol
3. 18.? mol 4. 9.3 mnl cataract 5. 18 mol
Expianation: 132/ P1 m “4/”:
2o.’r{31.o m mpsn m a: 13.7 males None that is l8.7 tricks of gas still in the.
tank. The ammmt used is murders... 28.0 "18.? a: 9.3 mills 02 013 10.13 points
Ammonium nitrate can decompose according
to the equation Nl‘LgNOﬁs} ma 91209;) + 2 H20(g) . How much GAS is produced by decompnailion
of 160 g nf swimming: nitrate at. “25°C and
86 kPa? 1 48.0 L
2. 57.6 L
3. 144 L com
4. 173 L 5. 14.5 L
Explanation:
‘71!“th z: 160 E N345“):
Fm (3619:.) Lance—13mm.
101.325 kPa ' T a W25°C 4 273.15 248.15 K
For the NH4N03, "mum. == {1508 WWW}
x 1 mo: NH4NO;
80.0434 1; NHgN 03
3* 199392 moi NH4N03 The qmstion asks fur the numbur of moles
of GAS; 129., moi ongO AM) H20: binIce {ﬁab2376}  l‘lomcwork 5 m Suwliﬂ'e  (53i20) '1' I. The average lancéic energies of molecqu
of samples of differwt idea? gases at the same
stimpemmm am the same. 2. The molecules of .1 gas move in unclaim
mg curved paths will! they collide with other
molecules 0: the walis of their containers, ac
cording to the lcineLicmulecular theory. cor
root 3. Ah ug‘lvcn bumperatuxc. according to Lhc
kinetic moiemﬂar theory, the avenge speed
of HF molecules is gramm than the average
speed of F2 molecules. 4. The molcwies of an ideal gas are rela
zively very far span an the average. 5. The avcmgc lime‘lic mmgjm of molecuiw
of samples of (liﬁ‘erenl. ideal gases at diﬁereut
twnpemtures are Meimt. Expiannhion:
Molecules of a gas move in unchmaging
straight name. 018 19.0 points
If equal weighm of ()2 and N; am placed in
separate containem of cquai whims at the
same temperature, which one cfthe following
is Luna? 1. Molecules in the oxygen ﬂask are moving
fazter on the average than the 0:395 in the
nitrogen ﬂask. 2. Both ﬂasks. contain the 5mm mxm’oer of
molecules. 3. The nitrogen moleculm have a greater
average kinetic energy. If. The prexum in the nicrugcu ﬂash in;
greater than that in the oxygen ﬂask. oar
reel. 5. None of these Explaxmﬁon: m ,  IT
UmP,Fxconstam,Uo< W E15; ‘4‘ Wm. x MWO.
m), J” E MWN:
We.
in
. I; m 1.5 ﬁg“ became the N9 molecules are moving
at an average kinetic speed 1.5 Limos faster
than 02 moleculw, and will therefore exert. a
higher {arc9mm. bmke (jpb231’6) w Humuwnrlc 5 —Sut.clﬂl”e— (53120) 6 3 mo! gas mm (1.99b'92molNHaN01) W 2 5.99675 moi gm:
’i‘hcidcal gas law lti PanRII;
nR '
Va P _ (5.996% mol gas) (0.08206 m} 0.848754 am
:4 (245.15 K) m 143.873 Laws 014 10.0 DOME; Sulfur dioxide reacts with wgen gas to pm
(lune mﬂfur trimdde. What volume of oxygen
gar. will: react with 15.0 I. of wifur dioxide if
bothgases are at. 101.3 “’3 and i25.°C? l. 3.75 L 2. 5.90 L 3. 1’50 L correct 4. 30.91. 5. 15.01”. Ex planation:
The balanced eqtmtion is 2809 +02 ma 2803 015 10.0 points
3.5 moles of oxygen molecula are mixeti with
22.53 ofAtgtm. What willbe ﬂieﬁnalvoimne
at STP? Contact any/er: 91.01.64 L.
Explanation: 016 10.0 points
A. 250 sampie of hydrogen was colv
lucked over water at. 20°C on a. day whm the M I
he— “‘1
f I...
rm"! A “u.
A"? V
3
9X3
\ “CPR. atmospheric pressure was 1'56 tort. The van
pm pressure of wan: untim ﬂame; conditions
is 17.54 ton. What: volume will the dry lxy‘
drogen (no water vapor pro59m) occupy at
STE"? 1. 0.226 ﬁbers correct
2. 0.385 lit/(Ks 3. 0.500 liters 1!. 20.0 lites 5. 0.105 liters Explanation:
V a 250 ml. : 0.25 L
T r: WC w" 293.15 K Pm“ u 756 tort
13mg m 17.54120" Pm: 1’ 17h: 4' Pllzo
PE; "—" Pam "* Pmo
w 756 tort — 17.54 fort _ 1 atm
m 'I'JSAO torr mo tor? 0.971658 nun
Applying the ideal gas iaw equation.
PV :1: n R 1"
PH: V
H T
+ (9.9?1658 aim) (0.25 L)
“ (0.08206 W) (293.15 K)
m 0.0100979 moi
Dry Hz at STE):
Hg, :7 0.0100970 moi Tau 0°C m 273.15 K
P I: l stun
Applying Lheiduai gas law inﬂation,
P V m HRT
nRT Vexm P _ (0.0100979 moi) (0.68206 2% ) 1 atm nu. # x 273.15 K
5 (2.225342 L 917 10.0 points
Whjd: statement is fake? ...
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This note was uploaded on 03/24/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
 Fall '07
 Fakhreddine/Lyon

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