|By Montydsw11 (Montydsw11) on Sunday, October 05, 2003 - 10:23 am: Edit|
Sorry for the double... err, triple post. Thanks in advance!!!
1. Propane, C3H8 is a hydrocarbon that is commonly used as fuel for cooking.
a)Write a balanced equation for the complete combustion of propane gas, which yields CO2(g) and H20(l)
b) Calculate the volume of air at 30 degrese Celcius and 1.00 atm needed to burn completely 10.0 grams of propane. Assume the air is 21% O2 by volume.
c)The heat of combustion of propane is -2200.1 kj/mol. Calculate the heat of formation, deltaHf, of propane given that deltaHf of H20(l) = -285.3j/gK and deltaHf oc COs(g) = -395.5kj/mol.
d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transfered to 8kg of water (specific heat = 4.184J/gk) calculate the increase in temperateure of water
2. A mixture of H2(g) and 2mL of H2O(l) are present in a .5 litre rigid container at 25 deg. C. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1146 mm Mercury. (The equilibrium vapor pressure of water is at 25 deg C us 24 mL mercury.) The mixture is sparked, and H2 and O2 react until one reactant is completely consumed.
a) identicy the reactant remaining and calculate the number of moles remaining
b) Calculate the total pressure in teh container at the conclusion of teh reaction if the final temp. is 90C. (The equilibrium vapor pressure of water at 90C is 526 mm mercury)
c) Calculate the number of moles of water present AS VAPOR in the container at 90C
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 10:25 am: Edit|
fuel + O2 ---> CO2 + H2O (hydrocarbon=fuel)
C3H8 (g) + 5O2 (g) ---> 3CO2 (g) + 4H2O
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 10:26 am: Edit|
While im working on this can you please look at my post (CHEM HELP) Thanks.
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 10:38 am: Edit|
10.0 g C3H8 * (1 mol / 44.096 g)*(5 O2 / 1 C3H8) = 1.13 mol of O2 needed to burn 10g of propane
T = 30 C ---> 30 +273 = 303 K
P = 1.00 atm
n = 1.13 mol
V = ???
(1.00 atm)(V)=(1.13 mol)(.0821 atm*L / mol*K)(303 K)
V = 28.1 L
.20*(VOLUME AIR) = 28.1 L
VOLUME AIR = 28.1/.20 L = 140.5 L (Use whatever s.f. your teacher wants, technically 20% has only 1 s.f. so it would be 100 L -- but that makes NO sense!)
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 10:47 am: Edit|
C3H8 (g) + 5O2 (g) ---> 3CO2 (g) + 4H2O /\H_comb = -2200.1 kJ/mol
heat of formation of propane = X
-2201.1 = (sum heats of formation of products) - (sum of heats of formation of reactants)
-2201.1 = ( (4*-285.3) + (3*-395.5) ) - ( (5*0) + X )
simplify & solve:
-2201.1 = -2327.7 -X
X = -2327.7 + 2201.1 = -126.6 kJ/mol
|By Montydsw11 (Montydsw11) on Sunday, October 05, 2003 - 11:20 am: Edit|
Thanks MO, ill take a look at urs a lil later... I am not that good at chem so I am not sure how useful I'll be, lol. Anyway, did you get #2?
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