Consider that calcium metal reacts with oxygen gas in the air to form calcium oxide. suppose we react 6.84 mol calcium with 4.00 mol oxygen gas. determine the number of moles of calcium oxide produced after the reaction is complete.

The balanced equation for the above reaction is as follows;
2Ca + O₂ ---> 2CaO
stoichiometry of Ca to O₂ is 2:1
we first need to find the limiting reactant 
number of Ca moles - 6.84 mol
number of O₂ moles - 4.00 mol
if Ca is the limiting reactant 
if 2 mol of Ca reacts with 1 mol of O₂
then 6.84 mol of Ca reacts with - 6.84 / 2 = 3.42 mol of O₂
this means that Ca is the limiting reactant and O₂ is in excess
therefore amount of CaO produced depends on amount of limiting reactant present 
stoichiometry of Ca to CaO is 2:2
number of moles of Ca reacted = number of CaO moles formed 
number of moles of CaO formed - 6.84 mol
answer is 6.84 mol

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