How many grams of \text{NaCl}NaClstart text, N, a, C, l, end text will be produced from 18.0 \text{ g}18.0 g18, point, 0, start text, space, g, end text of \text{Na}Nastart text, N, a, end text and 23.0 \text{ g}23.0 g23, point, 0, start text, space, g, end text of \text{Cl}_2Cl 2



start text, C, l, end text, start subscript, 2, end subscript?

18.7887 g of NaCl

Explanation:

The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?

Let us start by writing out the balanced equation of the reaction:

Na + Cl2 ---> NaCl2

1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.

mole = mass/molar mass

Therefore

18 g of Na = 18/23 = 0.7826 mole

23 g of Cl2 = 23/71 = 0.3239 mole

In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence

0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.

mass of 0.3239 mole NaCl = 0.3239 x 58 = 18.7887 g

Hence, 18.7887  grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.

answer; /// i do believe that the correct answer is (a) boiling

its first and third is your answer i did the qustion the fourth isnt right