1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?

Explanation

The balanced chemical equation for the reaction is :
`underset(48g)(2Mg)+underset(32g)(O_(2))rarrunderset(2(24+16)=80g)(2MgO)`
Step I. Determination of the limiting reactant
48 g of Mg react with oxygen = 32 g
1 g of Mg react with oxygen `= (32)/(48)=0.67` g
But the amount of oxygen actually available = 0.5 g
Since oxygen which is actually available is less than oxygen that is needed, therefore, oxygen is the limiting reactant
Step II. Calculation of the amount of MgO formed
According to the equation
32 g of oxygen react with Mg to form MgO = 80 g
0.5 g of oxygen react with Mg to form `MgO = (80)/(32) xx 0.5 g = 1.25 g`.