Question

Calculate the weight of ${{V}_{2}}{{O}_{5}}$ produced from 2 g of $VO$ and 5.75g of $F{{e}_{2}}{{O}_{3}}$.
\[VO+F{{e}_{2}}{{O}_{3}}\to FeO+{{V}_{2}}{{O}_{5}}\]

Answer 1

Hint: It is limiting reagent based numerical and in this first we have to find out the limiting reagent from the given reaction and the data according to stoichiometric calculations and then from this limiting reagent we can find the amount of product formed which is dependent on the limiting reagent. Now solve it.

Complete Solution :
This numerical is based on the concept of the limiting reagent. Many times, the reactions are carried out when the reactants are not present in the amounts as required by the balanced chemical reaction.
In such situations, one reactant is in excess over the other and the other reactant which is present in a lesser amount gets consumed and after that no reaction occurs and thus, it limits the amount of product formed and is, therefore, known as the limiting reagent.

- Now considering the statement :
The reaction is given as:
\[VO+F{{e}_{2}}{{O}_{3}}\to FeO+{{V}_{2}}{{O}_{5}}\]

- After balancing we get, the reaction as:
\[2VO+3F{{e}_{2}}{{O}_{3}}\to 6FeO+{{V}_{2}}{{O}_{5}}\] ---------------(1)

- Now, first of all we will find the limiting reagent on whose product form depends.
So, the molecular mass of $VO = 51 + 16 = 67g\text{ }mol{{e}^{-1}}$
Since, the balanced reaction involves two moles of $VO$, then
Molecular mass of 2 moles of $VO = 67\times 2 = 134 g$
Similarly, the molecular mass of $F{{e}_{2}}{{O}_{3}} = 56\times 2+16\times 3 = 112 + 48 = 160g\text{ }mol{{e}^{-1}}$

Since, the balanced reaction involves three moles of $F{{e}_{2}}{{O}_{3}}$, then
Molecular mass of 3 moles of $F{{e}_{2}}{{O}_{3}} = 160\times 3 = 480g$
Molecular mass of ${{V}_{2}}{{O}_{5}}=51\times 2 + 16\times 5 = 102 + 80 = 182g\text{ }mol{{e}^{-1}}$
Now from equation (1) , According to the stoichiometric calculations; we will find the limiting reagent as:
$\begin{align}
  & 134g\text{ }of\text{ }VO\text{ }requires = 480g\text{ }of\text{ }F{{e}_{2}}{{O}_{3}} \\
 & \,1g\text{ }of\text{ }VO\text{ }requires=\dfrac{480}{134}g\text{ }of\text{ }F{{e}_{2}}{{O}_{3}} \\
 & \text{ 2}g\text{ }of\text{ }VO\text{ }requires = \dfrac{480}{134}\times 2g\text{ }of\text{ }F{{e}_{2}}{{O}_{3}}\text{ } \\
 & \text{ =7}\text{.16 }g\text{ }of\text{ }F{{e}_{2}}{{O}_{3}}\text{ } \\
\end{align}$
But we have been given 5.75 g of $F{{e}_{2}}{{O}_{3}}$ in the statement.

- So, it means that $F{{e}_{2}}{{O}_{3}}$ is the limiting factor and the amount of product depends upon it.
From equation (1), according to stoichiometric calculations; the amount of product formed is;
$\begin{align}
  & \text{480g }of\text{ }F{{e}_{2}}{{O}_{3}}\text{ }requires = 182g\text{ }of\text{ }{{V}_{2}}{{O}_{5}} \\
 & \,1g\text{ }of\text{ }F{{e}_{2}}{{O}_{3}}\text{ }requires\text{ }=\dfrac{182}{480}g\text{ }of\text{ }{{V}_{2}}{{O}_{5}} \\
& \text{5}\text{.75}g\text{ }of\text{ }F{{e}_{2}}{{O}_{3}}requires=\dfrac{182}{480}\times 5.75g\text{ }of\text{ }{{V}_{2}}{{O}_{5}}\text{ } \\
 & \text{ =2}\text{.18 }g\text{ }of\text{ }{{V}_{2}}{{O}_{5}}\text{ } \\
\end{align}$
Therefore, the weight of ${{V}_{2}}{{O}_{5}}$ produced from 2 g of $VO$ and 5.75g of $F{{e}_{2}}{{O}_{3}}$ is 2.18 g.

Note: Always keep in mind that the reactions which involve the limiting factor i.e. the reagent which is fully consumed then in those reactions the weight of the product that is formed depends upon that limiting factor and not on the other reactant which is present in excess.