## JerJason 3 years ago Given a sample hydrate of .854g and an anhydrous salt (NiSo4) of .503g: (a) Calculate the mass of the water driven off (b) The Percent of water in the hydrate (c) The number of moles of water in the hydrate sample (d) The number of moles of the anhydrous salt (e) The molar ratio of anhydrous salt to water in the hydrate

For (a) I got the following: .854 g - .503g =.351g of H2O lost (b) .351g/.854g * 100 = 41.10% (c) $.351g of H20 * \frac{ 1mol H2O }{ 18.02 g H2O } = .0194 mol H20$ (d)$.503g x \frac{ 1mol NiSo4 }{ 237.83g }=.0021 mol NiSo4$ (e) $\frac{ .0194 mol H2O}{ .0021mol NiSo4 }= 9.238 mol; 1:9 ratio$