The reactant which is entirely consumed in a reaction is known as the limiting reagent. In the reaction $2A + 4B \to 3C + 4D$,when $5 \ mol$ of $A$ react with $6 \ mol$ of $B$,then:
$(i)$ Which is the limiting reagent?
$(ii)$ Calculate the amount of $C$ formed.
$(i)$ Which is the limiting reagent?
$(ii)$ Calculate the amount of $C$ formed.
(B) The given reaction is: $2A + 4B \to 3C + 4D$.
Given moles of $A = 5 \ mol$ and $B = 6 \ mol$.
To find the limiting reagent,we calculate the moles of product $C$ formed by each reactant:
For $A$: $2 \ mol$ of $A$ produces $3 \ mol$ of $C$. Therefore,$5 \ mol$ of $A$ produces $\frac{3}{2} \times 5 = 7.5 \ mol$ of $C$.
For $B$: $4 \ mol$ of $B$ produces $3 \ mol$ of $C$. Therefore,$6 \ mol$ of $B$ produces $\frac{3}{4} \times 6 = 4.5 \ mol$ of $C$.
Since reactant $B$ produces a smaller amount of product $C$,$B$ is the limiting reagent.
The amount of $C$ formed is $4.5 \ mol$.
Given moles of $A = 5 \ mol$ and $B = 6 \ mol$.
To find the limiting reagent,we calculate the moles of product $C$ formed by each reactant:
For $A$: $2 \ mol$ of $A$ produces $3 \ mol$ of $C$. Therefore,$5 \ mol$ of $A$ produces $\frac{3}{2} \times 5 = 7.5 \ mol$ of $C$.
For $B$: $4 \ mol$ of $B$ produces $3 \ mol$ of $C$. Therefore,$6 \ mol$ of $B$ produces $\frac{3}{4} \times 6 = 4.5 \ mol$ of $C$.
Since reactant $B$ produces a smaller amount of product $C$,$B$ is the limiting reagent.
The amount of $C$ formed is $4.5 \ mol$.
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