Question

How many litres of ${{O}_{2}}$ gas at NTP is needed to react completely with $27gm$ of $Al$?
(A) 5.6 litres
(B) 11.2 litres
(C) 22.4 litres
(D) 16.8 litres

Answer 1

Hint: Using the Avogadro’s law which states 1 mol of gas at NTP occupies 22.4 L of the gas, we can easily find the volume occupied by the oxygen atom at NTP but we have to find out the moles of oxygen from the moles of Al and moles of Al can be calculated by using the formula as; Moles of $Al=$ $\dfrac{given\text{ }mass}{molecular\text{ }mass}$.
Complete step by step answer:
This is based on Avogadro's law. So, let’s discuss Avogadro's law. This law states that any gas molecule at the standard conditions of temperature(i.e. $^{\circ }C$) and pressure(i.e. 1 atm) contains Avogadro number of particles and 1mol of the gas at STP or NTP occupies the volume of $22.4L$.
Now, considering the numerical;
First, we will find the moles of Al by the formula as;
Moles of $Al=$ $\dfrac{given\text{ }mass}{molecular\text{ }mass}$
As we know,
Given the mass of Al=27
The molecular mass of Al=27
Then, no of moles is;
Moles of $Al=$ $\dfrac{27}{27}$
$= 1$
Therefore, no of moles of Al=1
Now, when oxygen is made to react with aluminium, it forms aluminium oxide. The reaction is supposed to occur as:
\[4Al+3{{O}_{2}}\to 2A{{l}_{2}}{{O}_{3}}\]
So, from the reaction, we can see that;
4 moles of aluminium reacts with= 3moles of oxygen
Then,
1mole of aluminium will react with= $\dfrac{3}{4}$ moles of oxygen
$= 0.75 \text (moles of the oxygen)$
From Avogadro’s law, we know that,
1 mole of oxygen gas at NTP occupies= 22.4 L
Then,
$0.75$ moles of oxygen gas at NTP will occupy = ${22.} \times{ 0.75}$ $L$
$=16.8$ $L$
Thus, 16.8L of the oxygen gas at NTP is needed to react completely with 27 gm of Al.

Hence, option (D) is correct.

Note: Avogadro’s law is applicable only at the conditions of standard temperature and pressure and if the pressure and temperature are constant, then on increasing the amount of the gas, its volume also increases and vice-versa.