The balanced equation for the reaction is as follows 2Na + Cl₂ --> 2NaCl first we need to find the limiting reactant from the 2 reactants in the reaction stoichiometry of Na to Cl₂ is 2:1 number of Na moles - 133.0 g / 23 g/mol = 5.783 mol
at STP, molar volume is where 1 mol of any gas occupies a volume of 22.4 L if 22.4 L occupied by 1 mol then 33.0 L is occupied by 33.0 L / 22.4 L/mol = 1.47 mol therefore number of Cl₂ moles present - 1.47 mol
If Cl₂ is the limiting reactant 1 mol of Cl₂ reacts with 2 mol of NaCl therefore 1.47 mol of Cl₂ reacts with - 2 x 1.47 = 2.94 mol of Na
but 5.783 mol of Na is present which means Na is in excess and Cl₂ is the limiting reactant NaCl formed depends on amount of Cl₂ present stoichiometry of Cl₂ to NaCl is 1:2 number of NaCl moles formed - 2 x 1.47 mol = 2.94 mol of NaCl mass of NaCl formed - 2.94 mol x 58.5 g/mol = 172 g 172 g of NaCl is formed
