Answer: The volume of oxygen at STP is needed for complete given combustion is 1121.28 L.
Explanation:
The given reaction equation is as follows.
[tex]2C_{2}H_{18}(l) + 25O_{2}(g) \rightarrow 16CO_{2}(g) + 18H_{2}O(g)[/tex]
This shows that 2 moles of gasoline requires 25 moles of [tex]O_{2}[/tex]. Hence, moles of oxygen required to react with 4 moles of gasoline are calculated as follows.
[tex]\frac{25 mol O_{2}}{2 mol C_{8}H_{18}} \times 4 mol C_{8}H_{18}\\= 50 mol O_{2}[/tex]
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.
[tex]PV = nRT\\1 atm \times V = 50 mol \times 0.0821 L atm/mol K \times 273.15 K\\V = 1121.28 L[/tex]
Thus, we can conclude that volume of oxygen at STP is needed for complete given combustion is 1121.28 L.
