Question

The combustion of octane, C8H18, proceeds according to the reaction
2C8H18(l)+25O2(g)---->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?

Answer 1

297 mol Octane * 16 mol CO2 for each 2 mol Octane is equal to 2380 mol octane (since we only have three significant figures, we cannot express it with exact accuracy).  Now, we apply the ideal gas law:

V=nRT/P, or volume equals number of moles times the gas constant times temperature divided by pressure.  This means that V=2380*.082*(20+273)/.995=57500 liters.  That's a lot of carbon dioxide!

Answer 2

Answer:

57472.01 L

Explanation:

Given that:-

Moles of octane = 297 moles

According to the given reaction:-

$2C_8H_{18}_{(l)}+25O_2_{(g)}\rightarrow 16CO_2_{(g)}+18H_2O_{(l)}$

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