Question

Liquid benzene $\left( {{C_6}{H_6}} \right)$ burns in oxygen according to $2{C_6}{H_6}(I) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(l)$. How many liters of ${O_2}$ at STP are needed to complete the combustion of 39 g of liquid benzene ?
A. 74L
B. 11.2L
C. 22.4L
D. 84L

Answer 1

Hint: In this type of question first we will see the balancing of the reaction and after balancing the reaction we will calculate the amount of reactant in the balanced reaction. Now we can calculate the ratio of one reactant to another and finally according to the calculated ratio we can find the needed amount of reactant.

Complete step by step solution:
According to standard balanced equation of benzene with oxygen molecule (combustion of benzene)
We have, $2{C_6}{H_6}(I) + 15{O_2}(g) \to 12C{O_2} + 6{H_2}O$
So we can say that for the combustion of benzene 2 moles of benzene is required, 15 moles of oxygen molecule is required, and after combustion we get 12 moles of carbon dioxide and 6 moles of water.
Now we have molecular mass as follows
C=12 amu
H=1 amu
O=16 amu
Now we all know that the volume of one mole of gas at STP is 22.4 liter.
And for 2 moles of benzene 15 moles of oxygen molecule is required
So volume for 15 moles of oxygen will be $15 \times 22.4 = 336.0L$
$\therefore $molar mass of benzene (${C_6}{H_6}$) is 78 amu
So number of moles of benzene in 39 gram will be calculated as $\dfrac{{mass}}{{molar{\text{ }}mass}}$
So $\dfrac{{39}}{{78}} = \dfrac{1}{2} = 0.5$moles
Now we have for 2 moles of benzene 336 L of oxygen is required
Now for one moles of benzene $\dfrac{{336}}{2} = 168L$ of oxygen will be required
Similar for 0.5 moles of benzene required amount of oxygen will be $168 \times \dfrac{1}{2} = 84L$
Hence, we have calculated that 0.5 moles of gas will require 84 liter of oxygen molecules.

Therefore, option number D will be the correct option.

Note: In the above question for the given equation we have used a term called combustion. Combustion is the process of burning of a carbon containing compound in the presence of oxygen. So we have used the term combustion of benzene.