What is the greatest amount of magnesium oxide in moles that can be made with 6.3 mol Mg and 6.9 mol O2? Which reactant is limiting?How many moles of the excess reactant will be left afte the reaction

Answer 1

2 Mg+ O₂ ==> 2 MgO ... balanced equation

Limiting reactant = Mg since it takes 2 mols Mg for each mol of O2, so not enough Mg.

Theoretical yield of MgO = 6.3 mols Mg x 2 mol MgO / 2 mols = 6.3 mols MgO

Excess reactant is O2. How much is left over? Subtract mols used up from mols originally present.

moles used up = 6.3 mols Mg x 1 mol O2 / 2 mols Mg = 3.15 mols used up

moles left over = 6.9 mols - 3.15 mols = 3.8 mols left over (2 sig. figs.)