1.0 g of magnesium is burnt with 0.56 g O_ | vedclass

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(A) The balanced chemical reaction is: $2Mg + O_2 \rightarrow 2MgO$.Moles of $Mg = \frac{1.0 \ g}{24 \ g/mol} = 0.04167 \ mol$.Moles of $O_2 = \frac{0

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$Mg, 0.16 \ g$

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(A) The balanced chemical reaction is: $2Mg + O_2 ightarrow 2MgO$.Moles of $Mg = \frac{1.0 \ g}{24 \ g/mol} = 0.04167 \ mol$.Moles of $O_2 = \frac{0.56 \ g}{32 \ g/mol} = 0.0175 \ mol$.According to the stoichiometry,$2 \ mol$ of $Mg$ requires $1 \ mol$ of $O_2$.Therefore,$0.04167 \ mol$ of $Mg$ would require $\frac{0.04167}{2} = 0.020835 \ mol$ of $O_2$.Since we have only $0.0175 \ mol$ of $O_2$,$O_2$ is the limiting reagent.$Mg$ consumed $= 2 	imes 	ext{moles of } O_2 = 2 	imes 0.0175 = 0.035 \ mol$.$Mg$ left $= 0.04167 - 0.035 = 0.00667 \ mol$.Mass of $Mg$ left $= 0.00667 \ mol 	imes 24 \ g/mol = 0.16 \ g$.

$1.0 \ g$ of magnesium is burnt with $0.56 \ g \ O_2$ in a closed vessel. Which reactant is left in excess and how much? (At. $wt. \ Mg = 24, O = 16$)

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