A chemist heats 100.0 g of #FeSO_4 * 7H_2O# n a crucible to drive off the water. If all the water is driven off, what is the mass of the remaining salt?
1 Answer
#54.64# #"g"# anhydrate
With
Mols are interconvertible, as one mol of anything is the same number of particles as one mol of anything else.
#100.0 cancel("g FeSO"_4cdot7"H"_2"O") xx ("1 mol FeSO"_4cdot7"H"_2"O")/(278.02 cancel("g FeSO"_4cdot7"H"_2"O"))#
#=# #"0.3597 mols"#
For each
#0.3597 cancel("mols FeSO"_4cdot7"H"_2"O") xx "7 mols water"/cancel("1 mol FeSO"_4cdot7"H"_2"O")#
#=# #"2.518 mols water"#
Consequently, you have this mass of water by using its molar mass:
#2.518 cancel"mols water" xx ("18.015 g water")/cancel"mol water"#
#=# #"45.36 g water"#
So, your final mass should be