Consider that calcium metal reacts with oxygen gas in the air to form calcium oxide. suppose we react 7.43 mol calcium with 4.00 mol oxygen gas. determine the number of moles of calcium left over after the reaction is complete.

The balanced chemical equation for the above reaction is as follows;
2Ca + O₂ --> 2CaO
stoichiometry of Ca to O₂ is 2:1
this means that 2 mol of Ca reacts with 1 mol of O₂.
If O₂ is the limiting reactant, 
4 mol of O₂ should react with (4x2) - 8 mol of Ca
however only 7.43 mol of Ca is present. Therefore Ca is the limiting reactant.
7.43 mol of Ca reacts with - 7.43/2 = 3.715 mol of O₂
therefore there's excess O₂₂ remaining after the reaction
Since Ca is the limiting reactant, it is fully used up in the reaction and there is no Ca remaining after the reaction is completed.