I am practicing problems for my chemistry test and I am stuck on the following problem:
The value of $x$ in $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}$ can be found by determining the amount, in moles, of sulfate in the compound.
A $\pu{0.982 g}$ sample was dissolved in water and excess $\ce{BaCl2(aq)}$ was added. The precipitate of $\ce{BaSO4}$ was separated and dried and found to weigh $\pu{1.17 g}.$
Calculate the amount (in moles) of $\ce{BaSO4}$ in the $\pu{1.17 g}$ precipitate. And calculate the amount (in moles) of sulfate in the $\pu{0.982 g}$ sample of $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}.$
I was able to do the first part of the question:
$$n (\ce{BaSO4}) = \frac{\pu{1.17 g}}{\pu{233.38 g mol-1}} = \pu{0.005 mol}$$
But, I am unsure of the second part. Any help would be appreciated.