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
