The combustion of octane, c8h18, proceeds according to the reaction
2c8h18(l)+25o2(> 16co2(g)+18h2o(l)
if 297 mol of octane combusts, what volume of carbon dioxide is produced at 20.0 °c and 0.995 atm?

57472.01 L

Explanation:

Given that:-

Moles of octane = 297 moles

According to the given reaction:-

2 moles of octane on reaction produces 16 moles of carbon dioxide

1 mole of octane on reaction produces 16/2 moles of carbon dioxide

297 moles of octane on reaction produces 8*297 moles of carbon dioxide

Moles of carbon dioxide = 2376 moles

Given:  

Pressure = 0.995 atm

Temperature = 20.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (20.0 + 273.15) K = 293.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.995 atm × V = 2376 mol × 0.0821 L.atm/K.mol × 293.15 K  

⇒V = 57472.01 L