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