Question

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)?

Answer 1

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