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Is this correct or will this reaction not take place?

2007-04-19 19:06:11 · 7 answers · asked by pirate77 2 in Science & Mathematics Chemistry

7 answers

As these compounds are all water soluble, all that will exist in solution will be the corresponding ions. It is not possible to to tell if 2 moles of sodium nitrate and 1 mole of barium chloride where dissolved or 1 mole of barium nitrate and 2 moles of salt.

A chemist would not regard this as a chemical reaction. The arrow should be written as double ended to show the dynamic equilibrium that can take place.

2007-04-19 22:56:59 · answer #1 · answered by Anonymous · 0 0

I agree with ag_iitkg
This reaction won't take place .
To judge whether the reaction takes place or not, we should look at whether there are water,gases or solids produced.
If not, the reaction won't happen of course. Since Ba(NO3)2 and NaCl are all ions in the water, we couldn't see any changes and we shouldn't say it 's a reaction.

2007-04-19 19:36:17 · answer #2 · answered by Anonymous · 0 0

Ba(NO3)2+2NaCL---> BaCl2+ 2NaNO3 (put a 2 in front of NaCl [I'm assuming NaCL is a typo], and NaNO3)

2016-05-19 03:24:39 · answer #3 · answered by garnet 3 · 0 0

The balancing is correct and yes I believe this reaction will take place, anytime Na and Cl are allowed to react, they almost always form NaCl

2007-04-19 19:18:58 · answer #4 · answered by James B 1 · 0 0

The equation is properly balanced. As to will the reaction take place, that's why you're studying chemistry and I studied something else. A Handbook of Chemistry and Physics might give you a clue if your textbook does not.

2007-04-19 19:15:04 · answer #5 · answered by Helmut 7 · 0 0

Actually, all of them are salts of strong acids and strong bases. Also, they are ionic and well soluble in water. Thus, they will only dissociate into ions and not form any new compounds.

2007-04-19 19:20:51 · answer #6 · answered by ag_iitkgp 7 · 0 0

correct dy....
the reaction wil take place wa...

2007-04-20 02:17:35 · answer #7 · answered by FiRe BaLl 1 · 0 0

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