Question

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)

Answer 1

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.

`2C_(2)H_(18)(l) + 25O_(2)(g) \rightarrow 16CO_(2)(g) + 18H_(2)O(g)`

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.

`(25 mol O_(2))/(2 mol C_(2)H_(18)) * 4 mol C_(2)H_(18)\\= 50 mol O_(2)`

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.

`PV = nRT\\1 atm * V = 50 mol * 0.0821 L atm/mol K * 273.15 K\\V = 1121.28 L`

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