1well, heating them will distinguish
in general , how can carbonate vs. bicarbonate detection test be carried out??
-
UP 0 DOWN 0 0 7
7 Answers
6another quesn....
Colourless salt (A) ->(heating in presence of NaOH) -> (B) (gas )
A -> CaCl2 soln. -> white ppt. (C)
(B) gives white fumes with HCl
(C) decolourises acidified KMnO4
What is (A)? Explain reacns?
6another quesn...
A colourless salt (X) has 50% Na2SO3 and 50% H2O . What is the formula of (X) ? How much of SO2 at NTP is obtained when 2.52 g of (X) reacts with excess of dil.H2SO4?
1Q. A colourless salt (X) has 50% Na2SO3 and 50% H2O . What is the formula of (X) ? How much of SO2 at NTP is obtained when 2.52 g of (X) reacts with excess of dil.H2SO4?
Ans:
Molecular Mass = 126 + (7 x 18) = 252
2nd part :
Na2SO3.7H2O → Na2SO3 + 7H2O
Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2
Now applying unitary method,
252gm Na2SO3 gives 64gm SO2 ..
So, 2.52gm Na2SO3 gives 0.64gm SO2..
0.64 gm of SO2 will be formed..
1I think The anser to ur previous question is Ammonium Chloride.
When Ammonium salts r heated in presence of NaOH pungent smelling gas NH3 is released.
When a glass rod dipped in HCl is brought near da mouth of the test the white ammonia fumes intensify.
Wen CaCl2 is added to salt A the acid radical in the salt A is replaced by Cl.
hence the white ppt. is of NH4Cl.
AND chloride salts decolourise acidified KMno4.