calculate the volume of each product and the energy involved in the transformation for the complete combustion of 1,825.0 mL of butane (C4H10) gas and 18.00L of oxygen gas at 760.0 mmHg and 25.0C

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Using Documented Problem Solutions , calculate the volume of each product and the energy involved in the transformation for the complete combustion of 1,825.0 mL of butane (C4H10) gas and 18.00L of oxygen gas at 760.0 mmHg and 25.0C

SO ive found the written out solution with just answers but im trying to see how you get there and the Limiting reagents worked out and KJ Ill include a picture that shows the kind off showed out work im talking about. 

Predict reaction spontane ity and reversibility for he complete combution
of ethane
l010.0 me of Oxygen gas at 2s°cat 1.000 ATM,
gas
(CzHalai) when a50.0 ml of ethane is mixed with
1. Write and balance
2 CzHulg) +7Ozcg)
asc ACOzia) + 6HzOcg)
నం.w
10100me
2. Convert to moles
Oz:
LR
have
Czlto:
Since use
gases,
PV=nRt
n=PV
RT
n= 1.000 Arrill95002)
(08206LATuY 298k)
ni,0388 mnoles CzHo
n=.0413
and ID LR
Cor measured
%3D
guatiy)
mol K
N.04 moles
LR
N.04 moles
need
:17
2 mees
ACOzig)
3. Calculate AHo AS°
for balanced eg ant.
2n
mels
Ź so te Sig fig! 2 CZHL G) + 7O 24) →4C02lg) +
-6 HzOlg
-2855.4 KJ = AH÷(-1Yzmples)ES84.68 K%nfl)+O + Amolest-393.5 KS/D+ (6 mofls Y -24I.8 KS/hak)
%3D
TEA.36KJ
-1574.6 KJ
-1450.8 ES
93.0 A/k =ASo , "
1132.2 2/k
-459.0A/mlk -k35,00/k
for
AG° = AH°-TAS° = -2855,4 kJ-(298KY.0930kS/k)
854.89/k
4. Calculate AG° fa
balanced ea.amaănt
%D
%3D
2,833.1 KJ/-
7 moles Oz
21.7 kJ
5. Calcubte AG° for
LR amount
LR
.0413-2823.l KJ -17.0KJ
7 moles Oz
LR
so spant.
710KJ so imeversible,
I-way
Transcribed Image Text:Predict reaction spontane ity and reversibility for he complete combution of ethane l010.0 me of Oxygen gas at 2s°cat 1.000 ATM, gas (CzHalai) when a50.0 ml of ethane is mixed with 1. Write and balance 2 CzHulg) +7Ozcg) asc ACOzia) + 6HzOcg) నం.w 10100me 2. Convert to moles Oz: LR have Czlto: Since use gases, PV=nRt n=PV RT n= 1.000 Arrill95002) (08206LATuY 298k) ni,0388 mnoles CzHo n=.0413 and ID LR Cor measured %3D guatiy) mol K N.04 moles LR N.04 moles need :17 2 mees ACOzig) 3. Calculate AHo AS° for balanced eg ant. 2n mels Ź so te Sig fig! 2 CZHL G) + 7O 24) →4C02lg) + -6 HzOlg -2855.4 KJ = AH÷(-1Yzmples)ES84.68 K%nfl)+O + Amolest-393.5 KS/D+ (6 mofls Y -24I.8 KS/hak) %3D TEA.36KJ -1574.6 KJ -1450.8 ES 93.0 A/k =ASo , " 1132.2 2/k -459.0A/mlk -k35,00/k for AG° = AH°-TAS° = -2855,4 kJ-(298KY.0930kS/k) 854.89/k 4. Calculate AG° fa balanced ea.amaănt %D %3D 2,833.1 KJ/- 7 moles Oz 21.7 kJ 5. Calcubte AG° for LR amount LR .0413-2823.l KJ -17.0KJ 7 moles Oz LR so spant. 710KJ so imeversible, I-way
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