Suppose if I heat the products obtained on neutralising aqua regia with a mixture of NaOH and KOH in the ratio 1:3 (as it itself is a mixture of HCl and HNO3 in the ratio 1:3) to any temperature more (not less) than 100 degree Celsius (or 212 degree Fahrenheit) then can I get the 2 salts (NaNO3 & 3KCl) without 4H2O as it boils at that temperature?
According to the theory at ordinary temperature:
alkali + acid = salt + water. (reaction-exchange)
3HCl +HNO3+NaOH+3KOH=NaNO3 + 3KCl+ 4H 2 O
The acid and alkali in the solution exchange ions and mutually neutralize each other.
The essence of the neutralization reaction is the interaction of protons with hydroxide ions, resulting in the formation of weakly dissociating molecules: H++OH -⇔ H2O
Therefore, in this reaction, salts and water will be formed
But when acid and alkali are heated more (not less) than 100 degree Celsius , the water molecules are split off, it should form only the salt but not water (the salt in its anhydrous form):
aqua regia+mixture of NaOH and 3KOH=NaNO3+3KCl
Final updates to answer of question #164093 in General Chemistry-Line 3:(1)4H_2O,not 4H _2 O, and (2)OH^- not OH ^- as H^+ not H ^+ , Line 6:H_2O, not H2O, That's all
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I am telling you the final changes for this answer to update for the question, These are here: (1)Remove the spaces between (i)H and the subscript 2, and (ii)the subscript 2 and O in Line 3 of Para 1 (2)(i)Delete the space between OH and the minus sign just as there is no space between H and the plus sign,and (ii)Replace the ordinary 2 between H and O with a subscript 2 between both these elements in Line 6 of Para 1,this is what I want currently as the rest is all OK in this solution key,