In a reaction $A + B_2 \rightarrow AB_2$,identify the limiting reagent,if any,in the following reaction mixtures:
$(i)$ $300$ atoms of $A + 200$ molecules of $B_2$
$(ii)$ $2 \ mol \ A + 3 \ mol \ B_2$
$(iii)$ $100$ atoms of $A + 100$ molecules of $B_2$
$(iv)$ $5 \ mol \ A + 2.5 \ mol \ B_2$
$(v)$ $2.5 \ mol \ A + 5 \ mol \ B_2$
$(i)$ $300$ atoms of $A + 200$ molecules of $B_2$
$(ii)$ $2 \ mol \ A + 3 \ mol \ B_2$
$(iii)$ $100$ atoms of $A + 100$ molecules of $B_2$
$(iv)$ $5 \ mol \ A + 2.5 \ mol \ B_2$
$(v)$ $2.5 \ mol \ A + 5 \ mol \ B_2$
$A$ limiting reagent is the reactant that is completely consumed in a chemical reaction,thereby limiting the amount of product formed.
$(i)$ $1$ atom of $A$ reacts with $1$ molecule of $B_2$. Since $200$ molecules of $B_2$ are present,they will consume $200$ atoms of $A$. $100$ atoms of $A$ remain. Thus,$B_2$ is the limiting reagent.
$(ii)$ $1 \ mol$ of $A$ reacts with $1 \ mol$ of $B_2$. $2 \ mol$ of $A$ will consume $2 \ mol$ of $B_2$. $1 \ mol$ of $B_2$ remains. Thus,$A$ is the limiting reagent.
$(iii)$ $100$ atoms of $A$ react with $100$ molecules of $B_2$ in a $1:1$ ratio. No reactant is left. There is no limiting reagent.
$(iv)$ $5 \ mol$ of $A$ and $2.5 \ mol$ of $B_2$ are present. $2.5 \ mol$ of $B_2$ will consume $2.5 \ mol$ of $A$. $2.5 \ mol$ of $A$ remains. Thus,$B_2$ is the limiting reagent.
$(v)$ $2.5 \ mol$ of $A$ and $5 \ mol$ of $B_2$ are present. $2.5 \ mol$ of $A$ will consume $2.5 \ mol$ of $B_2$. $2.5 \ mol$ of $B_2$ remains. Thus,$A$ is the limiting reagent.
$(i)$ $1$ atom of $A$ reacts with $1$ molecule of $B_2$. Since $200$ molecules of $B_2$ are present,they will consume $200$ atoms of $A$. $100$ atoms of $A$ remain. Thus,$B_2$ is the limiting reagent.
$(ii)$ $1 \ mol$ of $A$ reacts with $1 \ mol$ of $B_2$. $2 \ mol$ of $A$ will consume $2 \ mol$ of $B_2$. $1 \ mol$ of $B_2$ remains. Thus,$A$ is the limiting reagent.
$(iii)$ $100$ atoms of $A$ react with $100$ molecules of $B_2$ in a $1:1$ ratio. No reactant is left. There is no limiting reagent.
$(iv)$ $5 \ mol$ of $A$ and $2.5 \ mol$ of $B_2$ are present. $2.5 \ mol$ of $B_2$ will consume $2.5 \ mol$ of $A$. $2.5 \ mol$ of $A$ remains. Thus,$B_2$ is the limiting reagent.
$(v)$ $2.5 \ mol$ of $A$ and $5 \ mol$ of $B_2$ are present. $2.5 \ mol$ of $A$ will consume $2.5 \ mol$ of $B_2$. $2.5 \ mol$ of $B_2$ remains. Thus,$A$ is the limiting reagent.
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