Step 1: Calculate molar mass of FeS.
\[
M(\text{FeS}) = M(\text{Fe}) + M(\text{S}) = 55.85 + 32.06 = 87.91 \, \text{g/mol}
\]
Step 2: Mass fraction of Fe in FeS.
\[
\text{Fraction of Fe} = \frac{55.85}{87.91} = 0.6353
\]
Step 3: Mass of Fe in 25 g sample.
\[
m_{\text{Fe}} = 25 \times 0.6353 = 15.883 \, \text{g}
\]
Step 4: Round off to 2 decimal places.
\[
m_{\text{Fe}} = 15.88 \, \text{g}
\]
Final Answer:
\[
\boxed{15.88 \, \text{g}}
\]
While doing Bayesian inference, consider estimating the posterior distribution of the model parameter (m), given data (d). Assume that Prior and Likelihood are proportional to Gaussian functions given by \[ {Prior} \propto \exp(-0.5(m - 1)^2) \] \[ {Likelihood} \propto \exp(-0.5(m - 3)^2) \] 
The mean of the posterior distribution is (Answer in integer)
Consider a medium of uniform resistivity with a pair of source and sink electrodes separated by a distance \( L \), as shown in the figure. The fraction of the input current \( (I) \) that flows horizontally \( (I_x) \) across the median plane between depths \( z_1 = \frac{L}{2} \) and \( z_2 = \frac{L\sqrt{3}}{2} \), is given by \( \frac{I_x}{I} = \frac{L}{\pi} \int_{z_1}^{z_2} \frac{dz}{(L^2/4 + z^2)} \). The value of \( \frac{I_x}{I} \) is equal to 