To determine which reactant is left in excess and how much, we need to analyze the reaction between magnesium (Mg) and oxygen (O$_2$).$\\ \\$The balanced chemical equation for the combustion of magnesium is:$\\ 2 \,$Mg$+$O$_2 \rightarrow 2 \,$MgO$\\ \\$From the equation, 2 moles of Mg react with 1 mole of O$_2$.$\\ \\$Calculate the moles of Mg:$\\$Given mass of Mg$= 1.0 \,$g$\\$Molar mass of Mg$= 24 \,$g/mol$\\$Moles of Mg$= \frac{1.0}{24} = 0.0417 \,$mol$\\ \\$Calculate the moles of O$_2$:$\\$Given mass of O$_2 = 0.56 \,$g$\\$Molar mass of O$_2 = 32 \,$g/mol$\\$Moles of O$_2 = \frac{0.56}{32} = 0.0175 \,$mol$\\ \\$According to the stoichiometry of the reaction, 2 moles of Mg require 1 mole of O$_2$.$\\$Therefore, 0.0417 moles of Mg would require:$\\ \frac{0.0417}{2} = 0.02085 \,$moles of O$_2 \\ \\$Since we only have 0.0175 moles of O$_2$, O$_2$ is the limiting reactant.$\\ \\$Calculate the amount of Mg that reacts:$\\$Using the stoichiometry, 1 mole of O$_2$ reacts with 2 moles of Mg.$\\$So, 0.0175 moles of O$_2$ will react with:$\\ 2 \times 0.0175 = 0.035 \,$moles of Mg$\\ \\$Convert moles of Mg reacted to grams:$\\$Mass of Mg reacted$= 0.035 \times 24 = 0.84 \,$g$\\ \\$Calculate the excess Mg:$\\$Initial mass of Mg$= 1.0 \,$g$\\$Mass of Mg reacted$= 0.84 \,$g$\\$Excess Mg$= 1.0 - 0.84 = 0.16 \,$g$\\ \\$Therefore, magnesium is left in excess, and the amount is 0.16 g. This corresponds to Option 1.