For the reaction 2na(s) + cl2(g) ➡️ 2nacl(s), how many grams of nacl (mm=58.5 g) could be produced from 133.0 of na(mm=23 g) and 33.0 l of cl2 (mm=71 g) at stp?

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 

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