Solveeit Logo
Login

Question

Question: Liquid benzene \(\left( {{C_6}{H_6}} \right)\) burns in oxygen according to \(2{C_6}{H_6}(I) + 15{O_...

Liquid benzene (C6H6)\left( {{C_6}{H_6}} \right) burns in oxygen according to 2C6H6(I)+15O2(g)→12CO2(g)+6H2O(l)2{C_6}{H_6}(I) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(l). How many liters of O2{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

Explanation

Solution

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, 2C6H6(I)+15O2(g)→12CO2+6H2O2{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 15×22.4=336.0L15 \times 22.4 = 336.0L
∴\therefore molar mass of benzene (C6H6{C_6}{H_6}) is 78 amu
So number of moles of benzene in 39 gram will be calculated as massmolar mass\dfrac{{mass}}{{molar{\text{ }}mass}}
So 3978=12=0.5\dfrac{{39}}{{78}} = \dfrac{1}{2} = 0.5moles
Now we have for 2 moles of benzene 336 L of oxygen is required
Now for one moles of benzene 3362=168L\dfrac{{336}}{2} = 168L of oxygen will be required
Similar for 0.5 moles of benzene required amount of oxygen will be 168×12=84L168 \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.