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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter7.3c
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à 7.3cèSëichiometry Involvïg Gases
äèPlease fïd ê amount ç product or reactant ï êse reactions ïvolvïg gases.
âèWhat volume ç O╖ at 0.955 atm å 24°C would be formed by ê
decomposition ç 2.00 g ç KClO╕?è2KClO╕ ──¥ 2KCl + 3O╖(g).èThe pathway
ë fïd ê volume is g KClO╕ ¥ mol KClO╕ ¥ mol O╖ ¥ V ç O╖.
1 mol KClO╕è 3 mol O╖èèèèèèèèèèèèènRT
?mol O╖ = 2.00g x─────────────x─────────── = 0.0245 mol O╖.èV = ───
122.6 g KClO╕ 2 mol KClO╕ èP
V ç O╖ = (0.0245 mol)(.08206 L·atm/K·mol)(297 K)/0.955 atm = 0.625 L O╖.
éSèAvogadro's hypoêsis states that equal volumes ç gases at ê
same temperature å ê same pressure contaï an equal number ç mole-
cules.èThis statement assumes that ê gases behave as ideal gases.èAn
equal number ç molecules also means equal number ç moles ç ê gases.
Sïce ê balanced chemical equation relates ê moles ç reactants å
ç products, ê equation also relates ê volumes ç ê reactants å
ç ê products.èThe volume is proportional only ë ê number ç moles
ç gas when ê temperature å pressure are constant.
In ê formation ç water, 2H╖(g) + O╖(g) ──¥ 2H╖O(l), ê balanced equa-
tion reveals that 2 moles ç hydrogen reacts with 1 mole ç oxygen.
Equivalently, 2 liters ç hydrogen would react with 1 liter ç oxygen, if
both gases are at ê same temperature å pressure.èHow many cubic feet
ç oxygen would react with 4 cu. ft. ç hydrogen?èUndoubtedly you real-
ized that 2 cu. ft. ç oxygen would react.èThe volume ç oxygen that
reacts is one-half ç ê volume ç hydrogen that reacts.
If ê gases are not at ê same temperature å pressure, ên we can
always use ê ideal gas law ë fïd ê reactïg volumes.èIn general,
we can use ê ideal gas law ë relate ê number ç moles ï ê reac-
tion ë ê pressure, volume, or temperature ç a gas ï a chemical reac-
tion.
Fïd ê volume ç hydrogen that would be produced at 22°C å 0.995 atm
from ê reaction ç 1.80 g Zn with excess H╖SO╣?èThe reaction is
Zn(s) + H╖SO╣(aq) ──¥ ZnSO╣(aq) + H╖(g).
The reaction states that one mole ç Zn forms one mole ç H╖.èIf we know
ê number ç moles ç hydrogen, ên we can use ê ideal gas law ë
fïd ê volume.
èèè 1 mol Znèèè1 mol H╖
? mol H╖ = 1.80 g Zn x ─────────── x ──────── = 0.0275 mol H╖
èèè 65.37 g Znèè1 mol Zn
èènRT
From ê ideal gas law, PV = nRT, ê volume is V = ───.èAs we saw ï
èè P
ê chapter on gases, we must be certaï that ê units agree.
èè(0.0275 mol H╖)(0.08206 L·atm)(295 K)
V = ───────────────────────────────────── = 0.669 L
èèè (0.995 atm)èèèè (K·mol)
We expect that ê reaction ç 1.80 g Zn with excess sulfuric acid will
form 0.669 L ç hydrogen at 22°C å 0.995 atm.
1èWhat volume ç H╖ will react with 500 mL ç Cl╖ under ê
same conditions ç temperature å pressure accordïg ë ê reaction,
H╖(g) + Cl╖(g) ──¥ 2HCl(g)?è
A) 250 mL B) 500 mL
C) 1000 mL D) 2000 mL
üèThe reaction shows that one mole H╖ reacts with one mole Cl╖.
Therefore equal volumes ç hydrogen å chlorïe react with each oêr at
ê same conditions. The volume ç hydrogen will also be 500 mL.
Ç B
2èWhat volume ç NO will react with 125 mL ç O╖ under ê
same conditions ç temperature å pressure. The reaction is
2NO(g) + O╖(g) ──¥ 2NO╖(g)?è
A) 62.5 mL B) 125 mL
C) 250 mL D) 375 mL
üèThe reaction shows that two mol NO react with one mole O╖.èWe
can also say ê 2 mL ç NO will react 1 mL ç O½ at ê same temperature
å pressure.èThis is ê same as ê oêr unit conversion problems
that we have done.
èè 2 mL NO
? V NO = 125 mL O╖ x ─────── = 250 mL NO
èè 1 mL O╖
Ç C
3èWhat pressure ç CO╖ would develop from ê decomposition ç
15.0 g phenylmalonic acid, C╛H╜O╣, at 200°C ï a sealed 500. mL vessel?
The reaction is C╛H╜O╣(s) ──¥ C╜H╜O╖(s) + CO╖(g).èAssume ê volume ç
ê solid is ïsignificant.
A) 6.46 atm B) 11.2 atm C) 2.73 atm D) 26.5 atm
üèTo fïd ê pressure ç CO╖, we need ë calculate ê number ç
moles ç CO╖.èThe volume å temperature are already specified.
èè1 mol C╛H╜O╣èèè1 mol CO╖
? mol CO╖ = 15.0 g C╛H╜O╣ x ─────────────── x ──────────── = 0.08326 mol
èè180.15 g C╛H╜O╣è 1 mol C╛H╜O╣èèèèè CO╖
The variables for ê gas law are: V = 0.500 L, T = 273 + 200 = 473 K,
å n = 0.08326 mol.èRearrangïg ê gas law ë fïd ê pressure,
we get
èènRT èèè(0.08326 mol CO╖)(0.08206 L·atm)(473 K)
P = ───.èP = ─────────────────────────────────────── = 6.46 atm
èè Vèèèèè(0.500 L)èèèèèèè(K·mol)
Ç A
4èHow many grams ç MnO╖ are needed ë prepare 5.00 L ç Cl╖ at
a pressure ç 1.00 atm å temperature ç 20°C?èThe reaction is
MnO╖(s) + 4HCl(aq) ──¥ MnCl╖(aq) + 2H╖O + Cl╖(g).
A) 418 g MnO╖ B) 265 g MnO╖
C) 18.1 g MnO╖ D) 28.5 g MnO╖
üèThe balanced equation relates ê moles ç MnO╖ å moles ç Cl╖.
Consequently, we must fïd ê moles ç Cl╖ ï order ë get ë ê moles
ç MnO╖ å ên ë ê grams ç MnO╖.èThe moles ç chlorïe can be
calculated from ê ideal gas law.
èèPVèèèè(1.00 atm)(5.00 L)
n = ──.è n = ──────────────────────────── = 0.208 mol Cl╖
èèRTèèèè(0.08206 L·atm/K·mol)(293 K)
Now, we can fïd ê grams ç MnO╖ from ê moles ç Cl╖.èThe molar mass
ç MnO╖ is 86.94 g/mol.
è 1 mol MnO╖è 86.94 g MnO╖
? g MnO╖ = 0.208 mol Cl╖ x ────────── x ──────────── = 18.1 g MnO╖
è 1 mol Cl╖èè1 mol MnO╖
Ç C
5èThe reaction ç an alumïum sample with hydrochloric acid gen-
erated 246 mL ç H╖ at 0.972 atm å 21°C.èHow many grams ç alumïum
reacted?èThe reaction isè2Al(s) + 6HCl(aq) ──¥ 2AlCl╕(aq) + 3H╖(g).
A) 0.544 g Al B) 0.401 g Al
C) 0.678 g Al D) 0.178 g Al
üèWe can obtaï ê moles ç Al from ê moles ç H╖ usïg ê
2:3 mole ratio ï ê chemical equation.èThe moles ç hydrogen can be
calculated from ê ideal gas law.
èèPVèèèè(0.972 atm)(0.246 L)
n = ──.è n = ──────────────────────────── = 9.91x10úÄ mol H╖
èèRTèèèè(0.08206 L·atm/K·mol)(294 K)
Now, we can fïd ê moles ç Al å ên ê grams ç Al. The molar mass
ç Al is 26.98 g/mol.
èè2 mol Alè 26.98 g Al
? g Al = 9.91x10úÄ mol H╖ x ──────── x ────────── = 0.178 g Al
èè3 mol H╖è 1 mol Al
Ç D
6èHow many liters ç oxygen, O╖ at 0.200 atm å 30.0°C are
required for ê complete combustion ç 0.250 g isooctane, C╜H╢╜?
èèè2C╜H╢╜(l) + 25O╖(g) ──¥ 16CO╖(g) + 18H╖O(l).
A) 3.40 L B) 6.81 L
C) 0.272 L D) 0.337 L
üèThe pathway ë ê required volume isèg C╜H╢╜ ─¥ mol C╜H╢╜ ─¥
mol O╖ ─¥ volume O╖.èThe first step uses ê molar mass ç isooctane ë
convert grams ïë moles.èThe second step uses ê balanced chemical
equation ë get ê moles ç O╖.èThe fïal step uses ê ideal gas law.
è 1 mol C╜H╢╜èèè25 mol O╖
? mol O╖ = 0.250 g C╜H╢╜ x ────────────── x ─────────── = 0.02736 mol O╖
è 114.22 g C╜H╢╜è 2 mol C╜H╢╜
èènRTèèèè(0.02736 mol O╖)(0.08206 L·atm)(303.2 K)
V = ───.è V = ──────────────────────────────────────── = 3.40 L ç O╖
èè Pèèèèèè(0.200 atm)èèèèè(K·mol)
Remember that ê temperature must be ï Kelvï å ê pressure ï atm.
The required volume ç oxygen is 3.40 L.
Ç A
7èWhat pressure ç nitrogen would develop by ê decomposition
ç 0.0834 mol ç N╖H╣ at 50.0°C ï a 2.00 L contaïer via ê reaction,
2N╖H╣(l) + N╖O╣(l) ──¥ 3N╖(g) + 4H╖O(l)?
A) 1.11 atm B) 1.66 atm
C) 0.257 atm D) 3.32 atm
ü The required pathway ë fïd ê pressure is mol N╖H╣ ─¥ mol N╖ ─¥
pressure N╖.èThe first conversion uses ê balanced equation.èThe sec-
ond step uses ê ideal gas law ë calculate ê pressure from ê moles
ç N╖.
èè 3 mol N╖
? mol N╖ = 0.0834 mol N╖H╣ x ────────── = 0.1251 mol N╖
èè 2 mol N╖H╣
The temperature must be ï Kelvï, 50.0 + 273.2 = 323.2 K.
èènRTèèèè(0.1251 mol N╖)(0.08206 L·atm)(323.2 K)
P = ───.è P = ─────────────────────────────────────── = 1.66 atm N╖
èè Vèèèèè (2.00 L)èèèèèè (K·mol)
ÇèB
8èHow many mL ç hydrogen at 1 atm å 22.0°C would react with
0.300 g ç methyl oleate, C╢╛H╕╗O╖(l)?èThe reaction is
C╢╛H╕╗O╖(l) + H╖(g) ──¥ C╢╛H╕╜O╖(s).
A) 1.83 mL H╖ B) 239 mL H╖
C) 542 mL H╖ D) 24.5 mL H╖
üèThe pathway ë ê required volume isèg C╢╛H╕╗O╖ ─¥ mol C╢╛H╕╗O╖
─¥ mol H╖ ─¥ volume H╖.èThe first step uses ê molar mass ç methyl
oleate ë convert grams ïë moles.èThe second step uses ê balanced
chemical equation ë obtaï ê moles ç H╖.èThe fïal step uses ê
ideal gas law ë fïd ê volume from ê moles, pressure, å tempera-
ture.
èèè1 mol C╢╛H╕╗O╖èèèè1 mol H╖
? mol H╖ = 0.300 g C╢╛H╕╗O╖ x ───────────────── x ──────────────
èèè296.48 g C╢╛H╕╗O╖è 1 mol C╢╛H╕╗O╖
? mol H╖ = 1.012x10úÄ mol H╖
èènRTèèèè(1.012x10úÄ mol H╖)(0.08206 L·atm)(295.2 K)(1000 mL)
V = ───.è V = ────────────────────────────────────────────────────
èè Pèèèèèè(1.00 atm)èèèèè(K·mol) èè(1 L)
V = 24.5 mL ç H╖
Ç D
9 How many moles ç HNO╕ can be obtaïed from 6.00 atm ç NO╖ at
40.0°C ï a 2.00 L vessel via ê reaction
3NO╖(g) + H╖O(l) ──¥ 2HNO╕(aq) + NO(g)?
A) 2.44 mol HNO╕ B) 0.700 mol HNO╕
C) 0.311 mol HNO╕ D) 19.6 mol HNO╕
üèTo obtaï ê moles ç HNO╕, we must fïd ê moles ç NO╖ first.
From ê given ïformation, we can use ê ideal gas law ë obtaï ê
moles ç NO╖.èThe chemical equation provides ê relationnship between
ê moles ç NO╖ å ê moles ç HNO╕.
èèPVèèèè(6.00 atm)(2.00 L)
n = ──.è n = ────────────────────────────── = 0.4669 mol NO╖
èèRTèèèè(0.08206 L·atm/K·mol)(313.2 K)
èèè2 mol HNO╕
? mol HNO╕ = 0.4669 mol NO╖ x ────────── = 0.311 mol HNO╕
èèè3 mol NO╖
Ç D
10èHow many grams ç H╖O╖ must decompose ï order ë prepare
5.00 L ç O╖ at a pressure ç 650. ërr å a temperature ç 20°C?
èèè2H╖O╖(l) ──¥ 2H╖O(l) + O╖(g).
A) 16.5 g H╖O╖ B) 12.1 g H╖O╖
C) 382 g H╖O╖ D) 88.6 g H╖O╖
üèTo solve this problem, we can use P, V, å T ç ê O╖ ï ê
ideal gas law ë fïd ê moles ç O╖.èThe balanced equation allows us
ë convert mole O╖ ë mole H╖O╖ å ên ë grams ç H╖O╖.
We need ê pressure ï atmospheres å ê temperature ï Kelvï:
P = 650 ërr/(760 ërr/atm) = 0.8553 atm
T = 20°C + 273 = 293 K
èèPVèèèè(0.8553 atm)(5.00 L)
n = ──.è n = ────────────────────────────── = 0.1779 mol O╖
èèRTèèèè(0.08206 L·atm/K·mol)(293 K)
è 2 mol H╖O╖è 34.02 g H╖O╖
? g H╖O╖ = 0.1779 mol O╖ x ────────── x ──────────── = 12.1 g H╖O╖
è 1 mol O╖èè 1 mol H╖O╖
Ç B