Reactants and Stoichiometry:
To determine which reactant is left in excess, we need to first determine the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2). The balanced equation is:

2Mg + O2 → 2MgO

From the balanced equation, we can see that 2 moles of magnesium reacts with 1 mole of oxygen to produce 2 moles of magnesium oxide. The molar mass of magnesium is 24.31 g/mol, and the molar mass of oxygen is 32.00 g/mol.

Calculating Moles:
Given that we have 1g of magnesium and 0.56g of oxygen, we can calculate the number of moles of each reactant using the formula:

moles = mass / molar mass

The moles of magnesium can be calculated as:
moles of Mg = 1g / 24.31 g/mol ≈ 0.041 moles

The moles of oxygen can be calculated as:
moles of O2 = 0.56g / 32.00 g/mol ≈ 0.0185 moles

Limiting Reactant:
To determine the limiting reactant, we compare the mole ratios of the reactants in the balanced equation. The ratio of magnesium to oxygen is 2:1.

From the calculations above, we can see that the moles of oxygen are less than half of the moles of magnesium. Therefore, oxygen is the limiting reactant.

Excess Reactant:
Since oxygen is the limiting reactant, there will be some magnesium left over after the reaction. To calculate the amount of excess magnesium, we need to determine the moles of magnesium oxide formed and subtract it from the moles of magnesium initially present.

From the balanced equation, we know that 2 moles of magnesium oxide are produced for every 2 moles of magnesium. Therefore, the moles of magnesium oxide formed are:

moles of MgO = (moles of Mg) / 2 = 0.041 moles / 2 = 0.0205 moles

The moles of excess magnesium can be calculated as:
moles of excess Mg = moles of Mg - moles of MgO = 0.041 moles - 0.0205 moles = 0.0205 moles

Conclusion:
In the reaction between 1g of magnesium and 0.56g of oxygen, oxygen is the limiting reactant, and magnesium is the excess reactant. There is approximately 0.0205 moles (or 0.498g) of excess magnesium remaining after the reaction.