Question

Calculate the volume of air containing $21\% $ by volume of oxygen at NTP required to convert $294c{m^3}$ of $S{O_2}$ into $S{O_3}$ under the same condition.
A) $200ml$
B) $300ml$
C) $500ml$
D) $700ml$

Answer 1

Hint: We have to know that the standard temperature and pressing factor are standard arrangements of conditions for trial estimations to be set up to permit correlations to be made between various arrangements of information. The most utilized norms are those of the International Union of Pure and Applied Chemistry (IUPAC) and the National Institute of Standards and Technology (NIST), albeit these are not all around acknowledged guidelines. Different associations have set up an assortment of elective definitions for their standard reference conditions.

Complete answer:
The condition of the response of sulfur dioxide and Oxygen to give sulfur trioxide, is given beneath:
$2S{O_2}\left( g \right) + {O_2}\left( g \right) \to S{O_3}\left( g \right)$
We can ascertain the moles of $S{O_2}$ and use it to discover the moles of ${O_2}$ needed by utilizing the condition.
We have $294c{m^3}$ of $S{O_2}$ . We can ascertain the moles of the $S{O_2}$ by utilization of Avogadro’s law.
Avagardo law expresses that 1 mole of any gas possesses precisely \[22.4L\] by volume at STP.
Having this as a main priority, let us convert \[22.4\] to liters.
$1000c{m^3} = 1L$
At that point,
$294c{m^3} = \dfrac{{294c{m^3}}}{{1000}} = 0.294L$
Convert \[0.294L\] to moles:
On the off chance that \[22.4L = 1mole\]
At that point,
\[0.294L = \dfrac{{1{\text{ }}mole \times 0.294}}{{22.4}} = 0.013125moles\]
Utilize the mole proportion in the condition to discover the moles of the oxygen required:
Mole proportion of Sulfur dioxide is \[2:1\]
Hence the moles of the oxygen utilized is
\[\dfrac{{0.013125}}{{2moles}} = 0.0065625moles\]
This is the quantity of moles of oxygen from the condition:
We will utilize a similar Avogadro’s law to discover the volume of oxygen from the moles
In the event that one mole involves \[22.4L\]
At that point \[0.0065625\] moles will involve,
\[\dfrac{{0.0065625moles \times 22.4}}{{1mole}} = 0.147Liter\]
The measure of oxygen spent in this response is \[0.147\] liters or\[147c{m^3}\].
The subsequent stage is to ascertain the measure of air that contains \[147c{m^3}\] of oxygen
On the off chance that \[21\% = 147c{m^3}\]
At that point,
\[100\% = \dfrac{{147 \times 100}}{{21}} = 700c{m^3}\]
Hence the volume of air required is \[700c{m^3}\].

Thus option D is correct.

Note:
We realize that, one mole of gas possesses \[22.4L\] at NTP one mole of \[{O_2}\] gas involves \[22.4L\]at NTP \[{O_2}\] contains two moles of oxygen particle. Subsequently two moles of oxygen atom possesses \[22.4L\] at NTP,
\[Mole = \dfrac{{Volume}}{{22.4}}\]
Therefore, one mole of an oxygen molecule contains \[\dfrac{{22.4}}{2} = 11.2L\] at NTP.