What is the method to be followed to prepare $\ce{Sn(NO3)2}$?
I tried to prepare $\ce{Sn(NO3)2}$ from reaction below:
$$\ce{SnCl2 + 2HNO3-> Sn(NO3)2 + 2HCl}$$
But $\ce{SnCl2}$ is not soluble. We are getting a turbid solution.
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Sign up to join this communityWhat is the method to be followed to prepare $\ce{Sn(NO3)2}$?
I tried to prepare $\ce{Sn(NO3)2}$ from reaction below:
$$\ce{SnCl2 + 2HNO3-> Sn(NO3)2 + 2HCl}$$
But $\ce{SnCl2}$ is not soluble. We are getting a turbid solution.
Tin nitrate can be prepared by mixing tin and nitric acid at various proportions.
Tin react with nitric acid to produce nitrate tin(II), nitrogen and water. Nitric acid - diluted solution. This reaction takes place slowly. Impurities: nitrogen(II) oxide (chemiday 1)
Tin react with nitric acid to produce nitrate tin(II), ammonium nitrate and water. Nitric acid - very dilute solution. This reaction takes place slowly.(chemiday 2)
Tin react with nitric acid to produce nitrate tin (II), oxide nitrogen (II) and water. Nitric acid - diluted solution.(chemiday 3)
A brief information on tin nitrate can be found here*
Stannous Nitrate is obtained in white leaflets having the composition $\ce{Sn(NO3)2.20H2O}$ by cooling to -20° C. a solution of stannous oxide in nitric acid of density 1.2. It is also formed by the action of highly diluted nitric acid on the metal, some of the acid thus suffering maximum reduction to ammonia:
$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$
According to C. H. H. Walker, tin dissolves in nitric acid forming stannous and stannic nitrates, the relative proportion of these two salts produced depending on the temperature and the strength of the acid; moreover, the yellowish white precipitate which separates when the somewhat concentrated acid is employed is said to be a hydrated, ill-defined, stannic nitrate.