If 4.00 moles of gasoline are burned according to the chemical reaction below, what volume of oxygen at STP is needed for complete
combustion?
2C2H18(1) + 2502(g) → 16CO2(g) + 18H2O(g)

The volume of oxygen at STP is needed for complete given combustion is 1121.28 L.

Explanation:

The given reaction equation is as follows.

This shows that 2 moles of gasoline requires 25 moles of . Hence, moles of oxygen required to react with 4 moles of gasoline are calculated as follows.

At STP, the pressure is 1 atm and temperature is 273.15 K. Therefore, using ideal gas equation the volume of oxygen is calculated.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

Thus, we can conclude that volume of oxygen at STP is needed for complete given combustion is 1121.28 L.

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >