In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
The answer to your question is Hubble’s law
1kg is about 2.2 pounds. So 4kg = 8.8 pounds. If that represent 19% of the total, here is the answer.
0.19*x=8.8
x=8.8/0.19
x=46.31 pounds
Use the universal gas formula
PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)
Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)
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