For the reaction #2Na(s) + Cl_2(g) -> 2NaCl(s)#, how many grams of #NaCl# could be produced from 103.0 g of Na and 13.0 L of #Cl_2# (at STP)?

1 Answer
Jan 28, 2017

Answer:

67,9 g

Explanation:

You have to search the limiting reagent between Na and #Cl_2#
You have # n= "mass"/"molecular mass" = (103,0 g)/(23,0 (g/(mol))) = 4,47 mol # of Na.
As 1 mole occupies 22,4 L at STP you have # n= (13,0L)/(22,4(L/(mol))) = 0,580# mol of #Cl_2#
The limiting reagent is #Cl_2#.
since from the balanced reaction you have 2 mol of NaCl from one of #Cl_2#, making the proportion from 0,580 mol of #Cl_2# you get 1,16 (mol) of NaCl that are
# 1,16 (mol) 58,5 g/(mol) = 67,9 g# of NaCl