Question

What volume of oxygen gas is required to react completely with 1.05 mol of benzene (CH) according to the following reaction at 0°C and 1 atm?benzene (C6H6) (l) + oxygen(g)—carbon dioxide (g) + water(g)________liters oxygen gas

Answer 1

We have the combustion reaction of benzene, the balanced equation of the reaction is:

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l)

We see that for every two moles of benzene, 15 moles of oxygen are needed, so the ratio O2 to C6H6 is 15/2.

Therefore, the moles of oxygen will be:

$\begin{gathered} molO_2=GivenmolC_6H_6\times\frac{15molO_2}{2molC_6H_6} \\ molO_2=1.05molC_6H_6\times\frac{15molO_2}{2molC_6H_6}=7.88molO_2 \end{gathered}$

To calculate the volume of oxygen, since oxygen is a gas, we will apply the ideal gas law which tells us:

$PV=nRT$

Where,

P is the pressure of the gas = 1atm

T is the temperature of the gas = 0°C =273.15K

n is the moles of the gas = 7.88molO2

R is a constant = 0.08206 atm.L/mol.K

We clear the volume, and replace the known data:

$\begin{gathered} V=\frac{nRT}{P}=\frac{7.88mol\times0.08206\frac{atm.L}{mol.K}\times273.15K}{1atm} \\ V=177L \end{gathered}$

Answer: The volume of oxygen gas required to react completely with 1.05 mol of benzene is 177 Liters oxygen gas