1.0 g of Mg is burnt with 0.28 g of O_2 in | vedclass

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(B) The balanced chemical equation is: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$.According to the stoichiometry,$48 \ g$ of $Mg$ reacts with $32 \ g$ of $

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$Mg, 0.58 \ g$

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(B) The balanced chemical equation is: $2Mg(s) + O_2(g) ightarrow 2MgO(s)$.According to the stoichiometry,$48 \ g$ of $Mg$ reacts with $32 \ g$ of $O_2$.For $1.0 \ g$ of $Mg$,the required amount of $O_2$ is $\frac{32 \ g \ O_2}{48 \ g \ Mg} 	imes 1.0 \ g \ Mg = 0.667 \ g \ O_2$.Since only $0.28 \ g$ of $O_2$ is available,$O_2$ is the limiting reagent and $Mg$ is in excess.Now,calculate the amount of $Mg$ consumed by $0.28 \ g$ of $O_2$: $\frac{48 \ g \ Mg}{32 \ g \ O_2} 	imes 0.28 \ g \ O_2 = 0.42 \ g \ Mg$.The amount of $Mg$ left in excess is $1.0 \ g - 0.42 \ g = 0.58 \ g$.

$1.0 \ g$ of $Mg$ is burnt with $0.28 \ g$ of $O_2$ in a closed vessel. Which reactant is left in excess and how much?

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