Determining sulfate content in Mohr's salt
If there is $\ce{0.00501 mol}$ of $\ce{BaSO4}$ in the precipitate, this means that there was $\pu{0.00501 mol}$ of sulfate in the Mohr's salt solution and $\pu{0.00501 mol}$ of sulfate in $\pu{0.982 g}$ of Mohr's salt. And there was one half, that is $\pu{0.0025 mol}$ of $\ce{Fe(NH4)2(SO4)2.xH2O}$ in $\pu{0.982 g}$ of Mohr's salt. This allows you to calculate the molar mass of the salt, and to deduce $x$.