Question

What weight of Al will be completely oxidized by 44.8 lit of oxygen at STP?
(A) 18 g
(B) 37.8 g
(C) 50.4 g
(D) 72 g

Answer 1

Hint: In order to find the weight of Aluminium which is oxidized by 44.8 litres of oxygen, we must know how to convert moles into litres. For that we have to first find the number moles of Aluminium which are oxidized by certain moles of oxygen.

Complete Solution :
Aluminium will undergo oxidation reaction in presence of oxygen to give Aluminium oxide. The reaction of oxidation of Aluminium is given below:
\[4Al + 3{O_2} \to 2A{l_2}{O_3}\]

From the above reaction, we can see that Aluminum and oxygen are in the ratio of \[4:3\].
We can say that 3 moles of oxygen will be required to oxidize 4 moles of Aluminium.
The number of moles of oxygen atom,
\[ = \dfrac{{44.8}}{{22.4}} = {\rm{2~moles~of~ }}{{\rm{O}}_{\rm{2}}}\]
Therefore, 2 moles of oxygen will be required to oxidize \[x\] moles of Aluminium.
\[\dfrac{x}{2} = \dfrac{4}{3}\]
\[x = \dfrac{{4 \times 2}}{3} = \dfrac{8}{3}\]
\[x = 2.66\]

Therefore, 2 moles of oxygen will be required to oxidize 2.66 moles of oxygen.
In order to find the weight of aluminium required to oxidize 44.8 litres of oxygen we have to take the product of the number of moles and molar mass of Aluminium. This is given by
\[{\rm{weight~of ~Al = number~of ~moles~of ~Al }} \times {\rm{ molar~mass~of~}}\]\[{\rm{weight~of~Al}} = 2.6627 = 71.82g\]

The 71.82 g can be approximated to 72 g
\[{\rm{weight~of~Al}} = 71.82 g \approx 72 g\]
72g of Aluminium will be completely oxidized by 44.8 lit of oxygen at STP.
So, the correct answer is “Option D”.

Additional information:
- A mole is the unit of measurement of the amount of substance present.
- One mole of substance or particles is equal to \[6.022 \times {10^{23}}\] particles.
- \[6.022 \times {10^{23}}\] is the Avogadro number.

Note: What STP means? STP which means standard temperature and pressure is an elementary concept. The temperature and pressure will be taken as a standard value, i.e. 760torr and 273K. We can also calculate the volume of a gas at STP by the use of the following formula
\[{V_{STP}} = V\dfrac{{{T_{STP}}}}{T} \times \dfrac{P}{{{P_{STP}}}}\]