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 dm^{3} CO2 2 dm^{3} C8H18

------------------ x ----------------- = 0.175 dm^{3} C8H18

1 16 dm^{3} 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)