The reaction between magnesium (Mg) and oxygen (O2) results in the formation of magnesium oxide (MgO) according to the following equation:
2Mg + O2 → 2MgO
The molar mass of Mg is 24.31 g/mol and the molar mass of O2 is 32.00 g/mol. Therefore, 1 g of Mg is equivalent to 0.041 moles and 1 g of O2 is equivalent to 0.03125 moles.
To determine the limiting reactant, we can calculate the moles of MgO produced from each reactant:
From Mg: 0.041 moles Mg × (2 moles MgO / 2 moles Mg) = 0.041 moles MgO
From O2: 0.03125 moles O2 × (2 moles MgO / 1 mole O2) = 0.0625 moles MgO
Since the amount of MgO produced from Mg is less than the amount produced from O2, Mg is the limiting reactant. Therefore, all the Mg will react to form MgO, leaving no Mg in the reaction vessel. The excess O2 will remain in the reaction vessel since it did not react completely with Mg. Thus, the answer is:
b) MgO only