
An element $A$ reacts with compound $B{O_3}$ to produce ${A_3}{O_4}$ and ${B_2}{O_3}$. The number of moles of ${A_3}{O_4}$ produced if one mole each of $A$ and $B{O_3}$ is reacted, is
1. $3$
2. $\dfrac{1}{3}$
3. $\dfrac{2}{3}$
4. $1$
Answer
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Hint: This question gives the knowledge about balanced chemical equations and a limiting reagent. The balanced chemical equation is defined as the equation in which the number of atoms in a molecule are equal in reactant and the product side.
Complete step-by-step answer:
The balanced chemical equation is defined as the equal number of atoms in the reactant side and the product side. The complete balanced equation for the reaction is as follows:
$9A + 8B{O_3} \to 3{A_3}{O_4} + 4{B_2}{O_3}$
Now, we will determine the number of moles of $A$ required as follows:
$8$ moles of $B{O_3}$ reacts with $9$ moles of $A$.
So, $1$ mole of $B{O_3}$ reacts with how many number of moles of $A$ is determined using unitary method as follows:
$ \Rightarrow \dfrac{{1 \times 9}}{8}$ moles of $A$.
Therefore, the number of moles of $A$ required is
$ \Rightarrow 1.125$
But, only 1 mole of $A$ is provided. Hence, $A$ is the limiting reagent.
Limiting reagent is defined as the reactant which is completely consumed during the chemical reaction. Limiting reagent determines the product of the reaction.
As we know, in the chemical reaction $9$ moles of $A$ produces $3$moles of \[{A_3}{O_4}\].
So, $1$ mole of $A$ produces how many number of moles of\[{A_3}{O_4}\] is determined using unitary method as follows:
$ \Rightarrow \dfrac{{1 \times 3}}{9}$ moles of\[{A_3}{O_4}\].
Therefore, the number of moles of ${A_3}{O_4}$ produced if one mole each of $A$ and $B{O_3}$ is reacted, is
$ \Rightarrow \dfrac{1}{3}$ moles of\[{A_3}{O_4}\].
Hence, the correct answer is $\left( 2 \right)$.
Note: You should have the complete knowledge about the balancing of chemical reactions. And also remember the limiting reagent determines the product of the chemical reactions. In balanced chemical reactions, the number of atoms in a molecule are equal in reactant and the product side.
Complete step-by-step answer:
The balanced chemical equation is defined as the equal number of atoms in the reactant side and the product side. The complete balanced equation for the reaction is as follows:
$9A + 8B{O_3} \to 3{A_3}{O_4} + 4{B_2}{O_3}$
Now, we will determine the number of moles of $A$ required as follows:
$8$ moles of $B{O_3}$ reacts with $9$ moles of $A$.
So, $1$ mole of $B{O_3}$ reacts with how many number of moles of $A$ is determined using unitary method as follows:
$ \Rightarrow \dfrac{{1 \times 9}}{8}$ moles of $A$.
Therefore, the number of moles of $A$ required is
$ \Rightarrow 1.125$
But, only 1 mole of $A$ is provided. Hence, $A$ is the limiting reagent.
Limiting reagent is defined as the reactant which is completely consumed during the chemical reaction. Limiting reagent determines the product of the reaction.
As we know, in the chemical reaction $9$ moles of $A$ produces $3$moles of \[{A_3}{O_4}\].
So, $1$ mole of $A$ produces how many number of moles of\[{A_3}{O_4}\] is determined using unitary method as follows:
$ \Rightarrow \dfrac{{1 \times 3}}{9}$ moles of\[{A_3}{O_4}\].
Therefore, the number of moles of ${A_3}{O_4}$ produced if one mole each of $A$ and $B{O_3}$ is reacted, is
$ \Rightarrow \dfrac{1}{3}$ moles of\[{A_3}{O_4}\].
Hence, the correct answer is $\left( 2 \right)$.
Note: You should have the complete knowledge about the balancing of chemical reactions. And also remember the limiting reagent determines the product of the chemical reactions. In balanced chemical reactions, the number of atoms in a molecule are equal in reactant and the product side.
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