the combination of sodium metal and chlorine gas yields solid sodium chloride?

Question

what is the equation for that?

Answer 1

2Na(s) + Cl2(g) ===> 2NaCl(s)

Answer 2

If you placed the proper ratio of sodium metal in a flask of chlorine gas you would see the metal burn. At the end of the reaction a white powder would remain at the bottom, and this is salt.

Answer 3

the sodium is react violently with water and chlorine is green coloured gas used in world war 1 and combination of these two form table salt

Answer 4

About as simple as you can get. Both elements are monovalent in the combination, so the equation is simply 2Na + Cl2 -> 2NaCl. Don't try this at home -- the reaction is highly exothermic. (Can you say BOOM?)

Answer 5

an person-friendly stoich situation. you merely could ascertain you regularly replace grams given to moles and paintings from there. step one million replace grams of limiting chemical to moles. 8.30 g Na / 23 g Na according to mole Na from the periodic table = 0.357 mol Na used. step 2 seem on the mole ratio of reactants and products on your balanced equation. 2 Na reactants factors 2 NaCl products, or 2 mol Na to 2 mol NaCl. which capacity 0.357 mol Na could provide you 0.357 mol NaCl. step 3 replace moles back to grams of things. 0.357 mol NaCl x fifty 8.5 g NaCl according to mol = 20.9 g NaCl **The theoretical yield is 20.9 g NaCl** % yield = (unquestionably yield / theoretical yield) x a hundred unquestionably yield is given as 19.5 g NaCl theoretical yield is calculated to be 20.9 g NaCl 19.5 / 20.9 = 0.933 x a hundred = ninety 3.3% yield **% yield is ninety 3.3%**

Answer 6

Cl+Na-> NaCl

Answer 7

2Na(s) + Cl2(g) = 2NaCl(s)