# 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)?

Jan 28, 2017

67,9 g

#### Explanation:

You have to search the limiting reagent between Na and $C {l}_{2}$
You have $n = \text{mass"/"molecular mass} = \frac{103 , 0 g}{23 , 0 \left(\frac{g}{m o l}\right)} = 4 , 47 m o l$ of Na.
As 1 mole occupies 22,4 L at STP you have $n = \frac{13 , 0 L}{22 , 4 \left(\frac{L}{m o l}\right)} = 0 , 580$ mol of $C {l}_{2}$
The limiting reagent is $C {l}_{2}$.
since from the balanced reaction you have 2 mol of NaCl from one of $C {l}_{2}$, making the proportion from 0,580 mol of $C {l}_{2}$ you get 1,16 (mol) of NaCl that are
$1 , 16 \left(m o l\right) 58 , 5 \frac{g}{m o l} = 67 , 9 g$ of NaCl