Question

In this reaction $2Al + \dfrac{3}{2}{O_2}$ formed $A{l_2}{O_3}$ , $9gm$ of Al will react with how many gm of ${O_2}$ ?

Answer 1

We know that Aluminium (Al) is a metal and when a metal reacts with oxygen, metal oxide will be formed. In the above reaction, when Aluminium is reacted with oxygen, it forms Aluminium oxide. First of all we will write the chemical equation for the above reaction and then deal in the number of moles of both reactants and products.

Complete Step By Step Answer:
We know that when a metal reacts with oxygen, metal oxide will be formed. In the above reaction, when Aluminium is reacted with oxygen, it forms Aluminium oxide.
The balanced chemical equation for the above reaction is given as follows:
$2Al + \dfrac{3}{2}{O_2} \to A{l_2}{O_3}$
Now, we will calculate the given mass (m)of Aluminium into number of moles (n) by dividing by the molar mass (M) of Aluminium as follows:
$n = \dfrac{m}{M}$
Molar mass of Aluminium is $27gmo{l^{ - 1}}.$
Given mass of Aluminium is $9g.$
$n = \dfrac{9}{{27}}$
$n = 0.334$
Now, we will use the stoichiometric relationships to determine the moles of ${O_2}$ .
Therefore, we can see from the equation that two moles of Aluminium (Al) reacts with $\dfrac{3}{2}$ moles of ${O_2}$ .
$0.334$ moles of Al reacts with:
$= \dfrac{{1.5}}{2} \times 0.334$
$= 0.25moles$
Mass of $0.25moles$ of oxygen molecule is given as:
$= 0.25 \times 32$
$= 8g$
Therefore, $9g$ of Aluminium will react with $8g$ of oxygen molecules.

Note:
To solve these types of questions, we should know the concepts of moles, relation between moles and molar mass. We should remember that generally metal oxides are basic in nature but Aluminium oxide is amphoteric oxide. This means that Aluminium oxide can react with both acid and base to form salt and water.