Question

Liquid benzene burns in oxygen according to:
${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\, + \,15{{\text{O}}_{\text{2}}}\, \to \,1{\text{2C}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{\text{6}}\,{{\text{H}}_{\text{2}}}{\text{O}}$
If density of liquid benzene is $0.88$ g/cc, what volume of ${{\text{O}}_{\text{2}}}$ at STP is needed to complete the combustion of $39$ cc of liquid benzene?
A. $11.2$ litre
B. $74$ litre
C. $0.074\,{{\text{m}}^3}$
D. $37\,{\text{d}}{{\text{m}}^3}$

Answer 1

By using density formula we will determine the mass of benzene then we will calculate the mole of benzene. We will calculate the volume of oxygen by using STP condition. Then we will compare the volume of oxygen with the mole of benzene to determine the volume of oxygen.

Complete step-by-step solution: The STP is known as standard temperature and pressure. The value of standard temperature in kelvin is $298\,{\text{K}}$. The value of the standard pressure in atm is ${\text{1}}\,{\text{atm}}$. One mole of a gas at $298\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$ volume.
The given reaction is as follows:
${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\, + \,15{{\text{O}}_{\text{2}}}\, \to \,1{\text{2C}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{\text{6}}\,{{\text{H}}_{\text{2}}}{\text{O}}$
According to the above reaction ${\text{15}}$ mole oxygen is reacting at STP.
${\text{1}}$ mole oxygen gas at STP = $22.4\,{\text{L}}$ volume
${\text{15}}$ mole oxygen gas at STP = $336\,{\text{L}}$ volume
Now, we will determine the mass of liquid benzene as follows:
${\text{density}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{volume}}}}$
On substituting $0.88$ g/cc for density and $39$cc for volume of liquid benzene,
$0.88\,{\text{g/cc}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{39}}\,{\text{cc}}}}$
$\Rightarrow {\text{mass}}\, = \,0.88\,{\text{g/cc}}\,\, \times {\text{39}}\,{\text{cc}}$
$\Rightarrow {\text{mass}}\, = \,34.32\,{\text{g}}$
So, the mass of liquid benzene is $34.32\,{\text{g}}$.
Now, we will determine the mole of liquid benzene as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
Molar mass of benzene is $78$g/mol.
On substituting $34.32\,{\text{g}}$ for mass and $78$g/mol for molar mass,
${\text{mole}}\,{\text{ = }}\,\dfrac{{34.32\,{\text{g}}}}{{{\text{78}}\,{\text{g/mol}}}}$
$\Rightarrow {\text{mole}}\,{\text{ = }}\,0.44$
So, the mole of liquid benzene is $0.44$ mol.
According to the given reaction ${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\, + \,15{{\text{O}}_{\text{2}}}\, \to \,1{\text{2C}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{\text{6}}\,{{\text{H}}_{\text{2}}}{\text{O}}$,
$2$ mole of liquid benzene is reacting with $336\,{\text{L}}$ volume of oxygen at STP so, $0.44$ mole of liquid benzene will react,
${\text{2mol }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\,{\text{ = }}\,{\text{336}}\,{\text{L}}\,{{\text{O}}_{\text{2}}}\,$
$\Rightarrow {\text{0}}{\text{.44}}\,{\text{mol }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\,{\text{ = }}\,{\text{74}}\,\,{\text{L}}\,{{\text{O}}_{\text{2}}}\,$
So, the volume of ${{\text{O}}_{\text{2}}}$ at STP is needed to complete the combustion of $39$ cc of liquid benzene is $74$ L.

Therefore, option (B) $74$L, is correct.

Note: We can also determine the moles of oxygen using in the reaction by comparing the stoichiometry. By stoichiometry comparison we will get the moles of oxygen then we will determine the volume of oxygen by using STP condition.