Question

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2A+4B\to 3C+4D,$ when 5 moles of A react with 6 moles of B, then (i) Which is the limiting reagent? (ii) Calculate the amount of C formed?

Answer 1

$2A+4B⟶3C+4D2A=5\text{moles}4B=6\text{moles}$

The limiting agent is $B$.

as $5$ $moles$ of $A$ requres $10$ $moles$ of $B$ but only $6$ moles are present.

Hence, $2$ $moles$ of $A=4$ $mole$ of $B$.

$x=6$ $moles$ of $B.$

$x=\frac{2\times 6}{4}=3$ $moles$.

Amount of $C$ consumed $=\frac{3\times 3}{2}=4.5$ $moles$ of $C.$