Question

In a closed vessel at STP, ${ 50L }$ of ${ CH }_{ 4 }$ is ignited with ${ 750L }$ of air (containing 20%). The number of moles of ${ O }_{ 2 }$ remaining in the vessel on cooling to room temperature is closest to:
A.${ 5.8 }$
B.${ 2.2 }$
C.${ 4.5 }$
D.${ 6.7 }$

Answer 1

Hint: One mole of a substance (atoms, molecules, ions, or particles) can be defined as the quantity of the substance in number in a mass of the substance equal to its atomic or molecular mass in grams. One mole of any substance contains ${ 6.022\times 10 }^{ 23 }$ particles (atoms, molecules, or ions), which is known as Avogadro number. When volumes of the gas are given, the number of moles can be calculated by taking the ratio of the given volumes to the volume of the gas at S.T.P i.s. 22.4 L

Complete answer:
As we know that 1 mole is equal to ${ 22.4 }$ liters at STP
Using formula,
${ Number\quad of\quad moles= }\dfrac { mass }{ molar\quad mass } $
So, ${ 50L }$ of ${ CH }_{ 4 }$ = $\dfrac { 50L }{ 22.4L/mol } $ = ${ 2.23 }$ moles
And 750L of air = $\dfrac { 750L }{ 22.4L/mol } $ = ${ 33.48 }$ moles
It is given that the air contains 20% ${ O }_{ 2 }$.
So, the number of moles of ${ O }_{ 2 }$ = ${ 33.48\times }\dfrac { 20 }{ 100 } $
                                  = ${ 6.697 }$ moles
As we know the following reaction will takes place;
${ CH }_{ 4 }{ +2O }_{ 2 }{ \rightarrow CO }_{ 2 }{ +2H }_{ 2 }{ O }$
Therefore, 2.23 moles of ${ CH }_{ 4 }$ will react with ${ 2\times 2.23 }$ = ${ 4.46 }$ moles of ${ O }_{ 2 }$.
Hence, the number of moles of ${ O }_{ 2 }$ remaining in the vessel on cooling to room temperature is = ${ 6.697- 4.46 }$ moles = ${ 2.2 }$ moles of ${ O }_{ 2 }$.

So, the correct option is B.

Additional Information:
We can communicate moles regarding sub-atomic mass, number of particles/particles and as far as volume.
1 mole of a substance which is pure has a mass (grams) which is equal to its molecular or atomic mass.
1 mole has the exact number of particles which is present in 12g of carbon-12 atoms.
1 mole of a gas occupies a volume of: ${ 22.4 }$ liters at S.T.P.

Note:
The possibility to make a mistake is that in this question the remaining number of moles are asked, so you have to subtract it from the total number of moles of ${ O }_{ 2 }$. If you do not subtract the number of moles from the total number of moles you will get the number of moles that have reacted not the number of moles that is remaining.