1.0 g of magnesium is burnt with 0.56 g of O2 in a closed vessel. Which reactant is left in excess and how much?

The balanced chemical reaction is shown below:

2Mg + O₂ → 2MgO

Moles of Mg = mass / molar mass = 1.0 g / 24.3 g/mol ≈ 0.041 mol
Moles of O₂ = mass / molar mass = 0.56 g / 32 g/mol ≈ 0.0175 mol

From the stoichiometry of the balanced equation, 2 moles of Mg react with 1 mole of O₂. Therefore, the mole ratio of Mg to O₂ is 2:1.

Let's determine the limiting reactant:

If all the O₂ reacts, it would require 2 * 0.0175 mol = 0.035 mol of Mg.
Since we have 0.041 mol of Mg, there is more Mg than needed to react with all the O₂. Therefore, O₂ is the limiting reactant.

Now let's calculate how much Mg is left in excess:

Mg reacted = 0.0175 mol O₂ * (2 mol Mg / 1 mol O₂) = 0.035 mol Mg
Mg remaining = 0.041 mol - 0.035 mol = 0.006 mol Mg
Mass of Mg remaining = 0.006 mol * 24.3 g/mol ≈ 0.146 g

The closest answer is Mg, 0.16 g. The slight difference is due to rounding in the molar mass calculations.