Question

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning completely 1.00 L of octane (C8H18, density = 0.7025 g/mL), yielding carbon dioxide and water?

Answer 1

Answer:

V = 8963 L

Explanation:

Our strategy here is to realize this problem involves stoichiometric calculation based on the balanced chemical equation for the combustion of octane:

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9 H₂O

From the density of octane we can obtain the number of moles:

D = m/V ⇒ m= D x V = 0.7025 g/mL x ( 1000 mL) = 702.5 g

MW octane  = 702.5 g/ 114.23 g/mol = 6.15 mol

Required  mol oxygen to react with octane:

6.15 mol octane x  25/2 mol O₂ / mol octane = 76.8 mol O₂

Now mol fraction is given by mol O₂ / total number mol air ⇒

n air = 76.8 mol O₂ / (0.2095 mol O₂ / mol air )  = 366.59 mol air

and from the ideal gas law we can compute the volume of air:

PV = nRT  ⇒ V = nRT/P

V = 366.59 mol air x 0.08205 Latm/Kmol x (25+ 273) K/ 1 atm

   = 8,963 Lts

Note we treat here  air as a compund which is allowed in combustion problems.

Answer 2

Volume of air will be "8968.49 L". A further explanation is below.

Given:

Density,

• 0.7025 g/mL

Volume,

• 1000 mL

Temperature,

• 25°C

Mass of octane,

= $Density\times Volume$

= $0.7025\times 1000$

= $702.5 \ g$

Moles of octane,

= $\frac{702.5}{114.23}$

= $6.149 \ moles$

The reaction,

• $2C_8 H_{18} (s)+25O_2 \rightarrow 16CO_2(g)+18 H_2O(g)$

Moles of oxygen necessary,

= $6.149\times \frac{25}{2}$

= $76.86 \ moles$

Now,

The moles of air,

= $\frac{PV}{RT}$

= $\frac{1.00\times V}{(0.0821\times 298)}$

= $0.0409 \ V$

Moles of oxygen will be:

= $0.2095\times 0.0490$

= $0.00857 \ V$

hence,

The volume of air will be:

→ $V = \frac{76.86}{0.00857}$

      $= 8968.49 \ L$

Thus the answer above is right.

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