Question

If rockets were fuelled with kerosene and liquid oxygen, what mass of oxygen would be required for every liter of kerosene? Assume kerosene to have the average composition ${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}}$ and density, $0.792{\text{ g/ml}}$.
A.$\;5.504{\text{ kg}}$
B.$\;2.752{\text{ kg}}$
C.$1.376{\text{ kg}}$
D.${\text{3}}{\text{.475 kg}}$

Answer 1

First, use the formula of density to find out the mass of kerosene. The formula of density is given as-
$\Rightarrow$ D=$\dfrac{{\text{m}}}{{\text{v}}}$ where m= mass and V is the volume.
Then we can easily calculate the number of moles of kerosene (composition is given) by using the formula- moles=$\dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$. Put the values in the given formula to find moles. Then calculate the number of moles of oxygen used for burning the obtained number of moles of kerosene. Once the number of moles of oxygen is calculated then using the formula of moles we can calculate the given mass of oxygen by using formula- Mass of oxygen=${\text{moles}} \times {\text{molecular mass}}$

Complete step by step answer:
Given, density of kerosene= $0.792{\text{ g/ml}}$
And the composition of kerosene = ${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}}$
We have to find the mass of oxygen required for every liter of kerosene.
Now we know that $1{\text{ L = 1000 ml}}$
Now, we know the formula of density is given as-
$\Rightarrow$ D=$\dfrac{{\text{m}}}{{\text{v}}}$ where m= mass and V is the volume.
Then on putting the values we get,
$\Rightarrow 0.792 = \dfrac{{\text{m}}}{{1000ml}}$
Then on adjusting, we get-
$\Rightarrow {\text{m}} = 0.792 \times 1000$
On solving, we get-
$\Rightarrow {\text{m}} = 792{\text{ gm/mol}}$
Now we know that moles=$\dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$ -- (i)
For kerosene, we know that the composition is${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}}$so we can calculate molecular mass.
$\Rightarrow {{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}} = \left( {14 \times 12} \right) + \left( {30 \times 1} \right)$
On solving we get-
Molecular mass=${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}} = 168 + 30 = 198$
Then on putting the given values in eq. (i) we get-
$\Rightarrow$ Moles of kerosene=$\dfrac{{792}}{{198}}$
On solving, we get-
$\Rightarrow$ Moles of kerosene=$4$
Now when kerosene fuel is burned then following reaction occurs-
${\text{2}}{{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}} + 43{{\text{O}}_2} \to 28{\text{CO + 30}}{{\text{H}}_2}{\text{O}}$
Here $2$ moles of kerosene react with $43$ moles of oxygen
So we can write-
$\Rightarrow 2{\text{ }}$ moles ${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}}$=$43$ moles ${{\text{O}}_2}$
But we found that the moles of kerosene =$4$
So $4$ moles ${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}}$=$43 \times 2$ moles of ${{\text{O}}_2}$
Then we get,
$4$ moles ${{\text{C}}_{{\text{14}}}}{{\text{H}}_{30}}$=$86$ moles of ${{\text{O}}_2}$
Now we know the values of moles of oxygen then using eq. (i) we can write-
Mass of oxygen=${\text{moles}} \times {\text{molecular mass}}$
We know that molecular mass of oxygen=$32$
So the required mass of oxygen= $86 \times 32$
On solving we get,
Required mass of oxygen for every liter of kerosene=$2752{\text{ gm}}$
Now we know that $1{\text{ kg = 1000 gm}}$
So $1{\text{ gm = }}\dfrac{{\text{1}}}{{1000}}{\text{kg}}$
Then $2752{\text{ gm}} = \dfrac{{2752}}{{1000}}{\text{kg}}$
On solving we get,
The required mass of oxygen= $2.752{\text{ kg}}$

Hence the correct answer is B.

Note:
Kerosene is a pale yellow or colourless volatile liquid which is a mixture of hydrocarbons. Besides being used as rocket fuel, it is used for the following purposes-
-It is used in kerosene lamps and domestic heaters for burning.
-It is also used as a solvent for insecticides and grease.
-It is commonly used in house-holds as a heating and cooking fuel.
-It is also used as a fuel for jets and aviation.