August 12, 2020, 04:56:45 PM
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### Topic: Gas Stoichiometry  (Read 297 times)

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#### kelluminati

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##### Gas Stoichiometry
« on: July 10, 2020, 08:22:11 PM »
Hi there!
I am confused on how to finish question 48 and 50. I've completed the steps that I think are necessary to start the problem but I'm not sure where to go next with it.
Also, I would appreciate if someone could confirm if I correctly completed question 49 and 51.
Thank you!

#### AWK

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##### Re: Gas Stoichiometry
« Reply #1 on: July 10, 2020, 10:05:31 PM »
48.
volume butane => moles butane => moles CO2 => volume CO2
49.
The answer should be in volume units
50.
Known moles of hydrogen, its volume, and pressure - calculation of temperature is simple
51.
AWK

#### MNIO

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##### Re: Gas Stoichiometry
« Reply #2 on: July 11, 2020, 12:28:34 AM »
problem 49.

you have this balanced equation
2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

we all know those coefficients are MOLE ratios right?
2 mol C8H18 reacts with 25 mol O2 to produce 16 mol CO2 + 18 mol H2O

BUT.. for gases at constant P and T, those coefficients are also VOLUME RATIOS.
2 L C8H18 + 25 L O2 --> 16 L CO2 + 18 L H2O

so.. if we assume all are gases at STP (which is a bad assumption but given the previous problem and the lack of density for liquid isooctane - we're going to have to do that), then you can easily write

1.4 dm3 CO2      2 dm3 C8H18
------------------ x ----------------- = 0.175 dm3 C8H18
1              16 dm3 CO2

**************
problem 51

your solution is correct but notice a couple of things
(1) when I calculated mole CO2, I carried 1 extra sig fig.
this is recommended for intermediate steps
(2) how I converted °C to K and kPa to atm in the last step? all inline

so you would enter this in your calculator as
0.04962 * .08206 * (273.15+12.3) / 107.4 * 101.325 =
or you could even do this
1.39 / 28.01 * 0.04962 * .08206 * 285.45 / 107.4 * 101.325 =
and avoid the intermediate calculation.

anyway.. good job.

balanced equation
2 CO + 1 O2 ---> 2 CO2

then converting 1.39g CO to moles CO2
1.39g CO       1 mol CO     2 mol CO2
------------- x ------------ x ------------- = 0.04962 mol CO
1          28.01g CO      2 mol CO

finally, assuming CO2 is ideal at those conditions
PV = nRT
v = nRT/P

0.04962 mol * 0.08206 Latm/molK * (273.15 + 12.3 K)
v = ---------------------------------------------------------------- = 1.10 L
107.4 kPa * (1atm / 101.325 kPa)