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