What is the theoretical yield of sodium chloride for the reaction of 36.0 g Na with 75.2 g Cl2 of the following 2NA+Cl2—> 2NaCl? I keep getting stuck with my conversions. I started with 36.0g NA (1 mol Na/22.99g Na) (1 mol Cl2/2mol Na) =0.782 mol Cl2….

The theoretical yield is 91.5g of ##NaCl##
Method 1
Start with the equation:
##2Na_((s))+Cl_(2(g))rarr2NaCl_((s))##
So 2 mole ##Na## reacts with 1 mole ##Cl_2## to give 2 moles##NaCl##
Convert moles to grams: ##A_r Na=23## ##A_rCl=35.5##
2×23 = 46g ##Na## reacts with 2x 35.5 = 71g ## Cl_2## to give 2x(23 +35.5 ) = 117g ##NaCl##
Use simple proportion to find how much would be made from 36g ##Na## by first working out what 1 g of ##Na## would give by dividing through by 46:
(46/46)g ##Na## reacts with (71/46) g ##Cl_2## to give (117/46)g ##NaCl##
So:
1g ##Na## reacts with 1.54g ##Cl_2## to give 2.54g ##NaCl##
So:
36g ##Na## reacts with 1.54×36= 55.44g ##Cl_2## to give 2.54×36=91.44g ##NaCl##
You can see that the chlorine is in excess so the theoretical yield is 91.5g of ##NaCl## to 1sf.
Method 2
Start with the balanced equation as above. Calculate the amount of NaCl that can form from each reactant.
36.0 g Na ##1 mol Na/22.99 g Na 2 mol NaCl/2 mol Na## = 1.566 mol NaCl (3 significant figures + 1 guard digit)
75.2 g Cl ##(1 mol Cl_2)/(70.91 g Cl_2) 2 mol NaCl/(1 mol Cl_2)## = 2.121 mol NaCl
The Na gives fewer moles of NaCl so Na is the limiting reactant.
The theoretical yield is
1.566 mol NaCl ##58.44 g NaCl/1 mol NaCl = 91.5 g NaCl##
Method 3
Divide the moles of each reactant by its coefficient in the balanced equation.
36.0 g Na ##1 mol Na/22.99 g Na = 1.566 mol Na## (3 significant figures + 1 guard digit)
75.2 g Cl ##(1 mol Cl_2)/(70.91 g Cl_2) = 2.121 mol Cl_2##
##2Na + Cl_2 2NaCl##
##1.566/2##; ##2.121/1##
0.7829; 2.121
NaCl gives fewer moles of reaction so NaCl is the limiting reactant.
So theoretical yield is
0.7829 mol reaction ##2 mol NaCl/1 mol reaction 58.44 g NaCl/1 mol NaCl## = 91.5 g NaCl