Question

sodium nitrate on reduction with zn in presence of naoh produces nh3. mass of sodium nitrate absorbing 1 mole of electron is

Answer 1

Final Answer : 10.625 g of NaNO3

Steps:
1) We have Reaction,
NaNO3 + Zn + NaOH ------> NH3 + H20 + other products

where Zn + NaOH acts as reducing agent.

Valency Factor : Oxidation state of Nitrogen in reactant - Oxidation stare of Nitrogen in Product
= 5-(-3) = 8

2) We know that,
By Equivalent Concept, 1 mole of electron will be absorbed by 1 equivalent of NaNO3 .

Mass of one equivalent of NaNO3
= Molar Mass / Valency Factor

$= \frac{23 + 14 + 16 \times 3}{8} \\ = \frac{85}{8} = 10.625g$
Hence, 1 mole of electrons will absorb 10.625g of NaNO3.

Answer 2

Answer:

10.625 g

Explanation:

eq. wt. = 85/8

= 10.625g

I hope this answer will help you....