Question

If 4.00 moles of gasoline are burned according to the chemical

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 $O_{2}$. Hence, moles of oxygen required to react with 4 moles of gasoline are calculated as follows.

$\frac{25 mol O_{2}}{2 mol C_{8}H_{18}} \times 4 mol C_{8}H_{18}\\= 50 mol O_{2}$

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.

$PV = nRT\\1 atm \times V = 50 mol \times 0.0821 L atm/mol K \times 273.15 K\\V = 1121.28 L$

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