To find the amount of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ present in the sample, we can follow the steps below, applying stoichiometry and the principle of equivalence in titration. Let's analyze each step to solve this problem:

When $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ is strongly heated, it decomposes, and the end product mentioned is $ \text{Mn}_3\text{O}_4 $. The chemical equation for the decomposition isn't provided directly, but it's crucial to know that we end up with $ \text{Mn}_3\text{O}_4 $ which indicates the reduction of the sulfate.

The $ \text{Mn}_3\text{O}_4 $ residue reacts with $ \text{FeSO}_4 $ in the presence of dilute $ \text{H}_2\text{SO}_4 $. Here, $ \text{Fe}^{2+} $ acts as a reducing agent.

The solution from step (ii) completely reacts with $ \text{KMnO}_4 $, using 50 ml of $ \text{KMnO}_4 $ solution.

Here, we learn that 25 ml of the $ \text{KMnO}_4 $ solution require 30 ml of 0.1 N $ \text{FeSO}_4 $ for complete reaction. This gives us a direct relationship to calculate the normality (and thus the molarity) of the $ \text{KMnO}_4 $ solution.

To find the normality of the $ \text{KMnO}_4 $ solution, we use:

$ V_1N_1 = V_2N_2 $

where,

• $ V_1 $ is the volume of $ \text{FeSO}_4 $ (30 ml),


• $ N_1 $ is the normality of $ \text{FeSO}_4 $ (0.1 N),


• $ V_2 $ is the volume of $ \text{KMnO}_4 $ (25 ml),


• $ N_2 $ is the normality of $ \text{KMnO}_4 $, which we will calculate.

$ 30 \times 0.1 = 25 \times N_2 $

$ N_2 = \frac{30 \times 0.1}{25} = \frac{3}{25} = 0.12\, \text{N} $

Now, knowing that 50 ml of $ \text{KMnO}_4 $ reacts with the $ \text{Mn}_3\text{O}_4 $ formed from the original $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $, and since 25 ml of $ \text{KMnO}_4 $ is 0.12 N, 50 ml of $ \text{KMnO}_4 $ is equivalent to 60 ml of 0.1 N $ \text{FeSO}_4 $. This equivalence helps us understand the amount of $ \text{Fe}^{2+} $ that would react with the $ \text{Mn}^{2+} $ from the original sample.

Given that 60 ml of 0.1 N $ \text{FeSO}_4 $ is used in total, and knowing that 1 mole of $ \text{MnSO}_4 $ reacts to eventually form $ \text{Mn}_3\text{O}_4 $ and is equivalent to the reaction of $ \text{Fe}^{2+} $ with $ \text{KMnO}_4 $, we can calculate the moles of $ \text{MnSO}_4 $ initially present.

$ \text{Moles of } \text{Fe}^{2+} = V \times N = 60 \times 0.1 \times 10^{-3} \text{ L} \times \text{N} = 6 \times 10^{-3} \text{ moles} $

Given the molar ratio of $ \text{Fe}^{2+} $ to $ \text{MnSO}_4 $ in the reactions (which can vary based on the specific reactions involved but typically could be 1:1 in redox reactions involving manganate and iron(II) ions in acidic solution), we can assume that the moles of $ \text{Fe}^{2+} $ directly give us the moles of $ \text{MnSO}_4 $ initially present because each $ \text{Fe}^{2+} $ ion reduces a $ \text{MnO}_4^- $ ion, and the $ \text{Mn}_3\text{O}_4 $ would be linked back to the $ \text{MnSO}_4 $ based on the stoichiometry not given.

Thus, the amount of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ present initially can be approximated if we know its molecular weight:

$ \text{MW of } \text{MnSO}_4\cdot4\text{H}_2\text{O} = 151.00 + (4 \times 18.015) = 151.00 + 72.06 = 223.06 \, \text{g/mol} $

Therefore, the mass of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ is:

$ \text{Mass} = 6 \times 10^{-3} \text{ moles} \times 223.06 \, \text{g/mol} = 1.33836 \, \text{g} $

In this way, we find that the sample originally contains approximately $ 1.34 $ g of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $. This calculation is based on an understanding of the stoichiometry of the reactions involved and the titration data. Some assumptions had to be made due to missing precise chemical equations, especially in the conversion between $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ to $ \text{Mn}_3\text{O}_4 $, and how it reacts with $ \text{KMnO}_4 $ and $ \text{FeSO}_4 $.