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2C8H18(I) + 25 O2 (g) = 16 CO2(g) + 18 H2O (g)If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm and the temperature is 35 C?

Sagot :

Answer

Volume of O2 = 1326 L

Explanation

Given:

2C8H18(I) + 25 O2 (g) = 16 CO2(g) + 18 H2O (g)

moles of gasoline = 4.00 mol

Pressure = 0.953 atm

Temperature = 35 C = 308 K

Required: volume of oxygen

Solution

Step 1: use stoichiometry to find the moles of oxygen

The molar ratio between gasoline and O2 is 2:25

Therefore the moles of o2 = 4.0 mol x (25/2) = 50.0 mol

Step 2: Calculate the volume using ideal gas law

PV = nRT

V = nRT/P

V = (50.0 x 0,08206 x 308)/0.953

V = 1326 L

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