60°C 100g apa...........48gFeSO4........148g solutie saturata
a.....................b.................468,66g
316,66..............152...............468,66g
10° 100g................22.................122g
316,66g..........152-x..............468,66 -x
22(468,66 - x) = 122(152 - x) 5155,26 - 11x = 9272 - 61x
50x = 4116,74 x = 82,3348g nFeSO4 = 82,3348/152 = 0,54 moli
mFeSO4·10H2O = 0,54·152 +10·0,54·18 = 179,53g