Identify the limiting reagent in the reaction $A + B_2 \to AB_2$ for 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) The balanced chemical equation is $A + B_2 \to AB_2$. According to the stoichiometry,$1$ atom of $A$ reacts with $1$ molecule of $B_2$.
$(i)$ $300$ atoms of $A$ require $300$ molecules of $B_2$. Since only $200$ molecules of $B_2$ are present,$B_2$ is the limiting reagent.
$(ii)$ $2 \ mol$ of $A$ require $2 \ mol$ of $B_2$. Since $3 \ mol$ of $B_2$ are present,$A$ is the limiting reagent.
$(iii)$ $100$ atoms of $A$ require $100$ molecules of $B_2$. Since both are present in stoichiometric amounts,there is no limiting reagent.
$(iv)$ $5 \ mol$ of $A$ require $5 \ mol$ of $B_2$. Since only $2.5 \ mol$ of $B_2$ are present,$B_2$ is the limiting reagent.
$(v)$ $2.5 \ mol$ of $A$ require $2.5 \ mol$ of $B_2$. Since $5 \ mol$ of $B_2$ are present,$A$ is the limiting reagent.
$(i)$ $300$ atoms of $A$ require $300$ molecules of $B_2$. Since only $200$ molecules of $B_2$ are present,$B_2$ is the limiting reagent.
$(ii)$ $2 \ mol$ of $A$ require $2 \ mol$ of $B_2$. Since $3 \ mol$ of $B_2$ are present,$A$ is the limiting reagent.
$(iii)$ $100$ atoms of $A$ require $100$ molecules of $B_2$. Since both are present in stoichiometric amounts,there is no limiting reagent.
$(iv)$ $5 \ mol$ of $A$ require $5 \ mol$ of $B_2$. Since only $2.5 \ mol$ of $B_2$ are present,$B_2$ is the limiting reagent.
$(v)$ $2.5 \ mol$ of $A$ require $2.5 \ mol$ of $B_2$. Since $5 \ mol$ of $B_2$ are present,$A$ is the limiting reagent.
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