Question

How many grams of oxygen is essentially required for complete combustion of 3 moles of butane gas?
A) 624 g
B) 312 g
C) 128 g
D) 64 g

Answer 1

Hint: Combustion reaction is a reaction in which a substance reacts with atmospheric oxygen. First write the balanced chemical reaction of butane reacting with oxygen. Examine the moles of butane and oxygen used in the reaction and then relate them.

Complete step by step answer:
As you know, combustion reaction is a chemical reaction in which a substance reacts with the atmospheric oxygen (${O_2}$) to produce carbon dioxide and water. Therefore, balanced combustion reaction of butane (${C_4}{H_{10}}$) will be as follows:
${C_4}{H_{10}} + \dfrac{{13}}{2}{O_2} \to 4C{O_2} + 5{H_2}O$

You can observe in the above reaction that, $\dfrac{{13}}{2}$ moles of oxygen are required for combustion of 1 mole of butane.
Therefore, going by unitary method, for complete combustion of 3 moles of butane, 19.5 $\left( {\dfrac{{13}}{2} \times 3} \right)$ moles of oxygen would be required.
Now, we know, number of moles = $\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
Molar mass of oxygen is 16 g/mol
We need to find a given mass which is the mass of oxygen required when the number of moles is equal to 19.5.
Thus, given mass of oxygen or mass of oxygen required = number of moles of oxygen $ \times $ molar mass of oxygen
Mass of oxygen required = 19.5 moles $ \times $ 16 g/mol = 312 g.
Hence, 312 grams of oxygen is essentially required for complete combustion of 3 moles of butane gas.
So, the correct answer is “Option B”.

Note: Combustion process is also known as burning. It is a high temperature exothermic redox chemical reaction between a substance or fuel (acts as reductant) and an oxidant (usually atmospheric oxygen). So, in the combustion reaction of butane, butane is the reductant and obviously, oxygen is the oxidant.