Limiting Reactants/Chemistry Problem!

Determine the weight of MgSO4*6H2O (s) that would be formed by the reaction of 90.34 g of MgSO4 (s) and 73.19 g of H2O.

For the weight of MgSO4*6H2O I got 497.14 g. Is that correct?

(90.34 g MgSO4) / (120.3683 g/mol) = 0.750529832 moles MgSO4
(73.19 g H2O) / (18.0153 g/mol) = 4.06265785 moles H2O

0.750529832 moles of MgSO4 would absorb completely 6 x 0.750529832 = 4.50317899 moles H2O, but there isn't that much H2O present, so H2O is the limiting entity.

(4.06265785 moles H2O) x (1/6) x (228.4602 g/mol) = 154.7 g MgSO4*6H2O