Question

2 g FeSO4 is completely oxidised by acidic
0.05 M KMnO, solution then what will be volume
of KMnO, required
(Fe = 56, S = 32, 0 = 16)
(1) 0.10 L
(2) 0.05 L
(3) 0.30 L
(4) 0.40 L​

Answer 1

Answer:

feso4 +kmno4 -- fe(+3)  + mn(2)

nfactor for kmno4 is 5

then gramequivalance of feso4 is equal to gram equivalence of kmno4

W/E=N*V      N=MOLARITY *nfactor  

2/152=0.05*5*v

v=0.05

Answer 2

Answer:

The volume of 0.05M KMnO₄ solution required to oxidize 2g of FeSO₄ will be equal to 0.05L.

Therefore, the option (2) is correct.

Explanation:

The balanced chemical reaction:-

$10FeSO_{4} \;\; +2KMnO_{4}\;+ 8H_{2}SO_{4}$  $\longrightarrow\; K_{2}SO_{4} \; + 2MnSO_{4}\; + \; 5Fe_{2}(SO_{4})_{3}\;$

                                                                                                   $+\; 8H_{2}O$

From the above chemical reaction, we can say that the 10 moles of  FeSO₄ oxidised 2 moles of KMnO₄ .

Molecular mass of FeSO₄  $=151.8gmol^{-1}$

Molecular mass of KMnO₄  $=158gmol^{-1}$

$10*151.8g$ of FeSO₄  will oxidise KMnO₄ $=2*158g$

2g of FeSO₄ will oxidise KMnO₄ $=\frac{(2)*(158)*(2)}{(10)*(151.8)}=0.416g$

So, amount of  KMnO₄ will be required = 0.416g

Given, Molarity of KMnO₄ solution = 0.05M

$Molarity = \frac{Mass\;\; of \;KMnO_{4}}{(Molar\; mass)(Volume)}$

$0.05=\frac{(0.416)}{(158)*V}$

Volume of KMnO₄ solution $=\frac{0.416}{(158)*(0.05)}=0.05L$

Therefore, 0.05 liter of 0.05M KMnO₄ solution is required to oxidide 2g of FeSO₄.