Question

Hydroxyl amine reduces iron (III) according to following equation
N2OH+ Fe (SO4)3= N2(g) + H2O+FeSO4+H2SO4
Which statement is correct
1) n-factor for Hydroxyl amine is 2
2) equivalent weight of Fe (SO), is M/2
3) 6 meq of Fe (SO.), is contained in 3 millimoles of ferrie sulphate
4) all of these​

Answer 1

Answer:

Hydroxyl amine reduces iron (III) according to following equation

N2OH+ Fe (SO4)3= N2(g) + H2O+FeSO4+H2SO4

Which statement is correct

answer :-2) equivalent weight of Fe (SO), is M/2

Plz Mark me as BRAINLIEST

Answer 2

Answer:

For reaction,

$NH_{2}OH+Fe_{2}(SO_{4})_{3}$ → $N_{2} (g)+H_{2}O+ FeSO_{4} +H_{2}SO_{4}$

Statement 2) equivalent weight of $Fe_{2} (SO_{4} )_{3}$, is M/2 correct.

Explanation:

• For a reaction,

$NH_{2} OH+Fe_{2} (SO_{4} )_{3}$ → $N_2}(g) +H_{2} O+FeSO_{4} +H_{2} SO_{4}$,

• n factor or $n_{f}$ (oxidation number) can be calculated as 1×3=3 and 2×1=2

• The equivalent weight of  $Fe_{2}(SO_4}) _{3}$ can be calculated as - M/2

• The equivalent weight of $Fe_{2} (SO_{4} _{})_{3}$ = $\frac{M}{2}$

⇒ Miliequivalent wt. of $Fe_{2}(SO_{4}_{})_{3}$ = Milimoles of $Fe_{2}(SO_{4})_{3}$ × 2

Hence $\frac{Miliequivalent wt. of Fe_{2} (SO_{4} )_{3} }{milimoles of Fe_{2} (SO_{4} )_{3} }$ = $\frac{2}{1}$

Therefore statement 2) saying the equivalent weight $Fe_{2} (SO_{4})_{3}$ , is M/2 is correct.