Introduction:
In this reaction, 1 g of magnesium reacts with 0.56 g of oxygen in a closed vessel. We need to determine the reactant that is left in excess after the reaction.

Calculation:
To determine the excess reactant, we need to compare the stoichiometry of the balanced chemical equation with the given amounts of reactants.

The balanced chemical equation for the reaction between magnesium and oxygen is:
2Mg + O2 -> 2MgO

According to the equation, 2 moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide.

Step 1:
First, we need to calculate the number of moles of each reactant.

The molar mass of magnesium (Mg) is 24.31 g/mol, so the number of moles of magnesium can be calculated as:
moles of Mg = mass of Mg / molar mass of Mg = 1 g / 24.31 g/mol = 0.041 moles of Mg

The molar mass of oxygen (O2) is 32 g/mol, so the number of moles of oxygen can be calculated as:
moles of O2 = mass of O2 / molar mass of O2 = 0.56 g / 32 g/mol = 0.0175 moles of O2

Step 2:
Next, we compare the moles of magnesium and oxygen to determine the limiting reactant.

From the balanced equation, we can see that 2 moles of magnesium react with 1 mole of oxygen. This means that 0.041 moles of magnesium would require 0.0205 moles of oxygen for complete reaction.

Since we have 0.0175 moles of oxygen, which is less than the required amount, oxygen is the limiting reactant.

Step 3:
To determine the excess reactant, we subtract the moles of the limiting reactant from the moles of the other reactant.

Excess moles of magnesium = moles of magnesium - moles required for complete reaction
= 0.041 moles - 0.0205 moles = 0.0205 moles

Step 4:
Finally, we calculate the mass of the excess reactant using its molar mass.

Mass of excess magnesium = excess moles of magnesium × molar mass of magnesium
= 0.0205 moles × 24.31 g/mol = 0.49855 g

Conclusion:
In the given reaction, 0.49855 g of magnesium is left in excess after the complete reaction with 0.56 g of oxygen.