1 g of Mg is burnt in a closed vessel which contains 0.5 g of dioxygen. what is the limiting reagent? What is the amount of Mgo formed in the reaction?

Burning of Magnesium:
2Mg + O₂ $\to$ 2MgO
​Mass of Mg = 1g
Molar Mass of Mg = 24g
​Moles of Mg = $\frac{1}{24}= 0.041mol$

​Mass of O₂ = 0.5g
Molar mass of O₂ = 32g
​Moles of O₂ = $\frac{0.5}{32} = 0.015mol$

1mol of O₂ reacts with 2 moles of Mg
Therefore, 0.015mol will react with 2$\times$0.015mol of Mg i.e., 0.03mol of Mg.
Since we have 0.041mol of Mg, so (0.041-0.03)mol = 0.011mol of Mg will be left over and Oxygen will be limiting reagent.

Amount of MgO formed in the reaction:
1 mol of O2 gives 2 moles of MgO, so 0.015mol of O₂ will give 0.03mol og MgO.
Molar mass of MgO = 40.3g
Mass of MgO formed = molar mass $\times$ moles = 40.3$\times$0.03 = 1.209g