Question
Aluminum sulfate exist as the hydrate Al2(SO4)3 all the numbers are subscriptssorry. If the water of hydration is removed from a 10.96 gramsample of Al2(SO4)3 times XH20, the remaining anhydrous Al2(SO4)3would have a mass of 5.63 grams. Caculate X, the x before theh20 Need this asap kinda waited till lasst minute, please usestoichometry so i can understand how u get answer. Aluminum sulfate exist as the hydrate Al2(SO4)3 all the numbers are subscriptssorry. If the water of hydration is removed from a 10.96 gramsample of Al2(SO4)3 times XH20, the remaining anhydrous Al2(SO4)3would have a mass of 5.63 grams. Caculate X, the x before theh20 Need this asap kinda waited till lasst minute, please usestoichometry so i can understand how u get answer.
Explanation / Answer
Given mass of the sample , m = 10.96 g mass of the anhydrous sample , m' = 5.63 g mass of the water , m" = m - m" = 10.96 - 5.63 =5.33 g No . of moles of water , n ' = mass / Molar mass = m" /18 = 5.33 / 18 = 0.296 moles No . of moles of anhydrous salt , n = m' / Molar mass of Al2 (SO4 ) 3 = 5.63 / 342 = 0.01646 ration to the no . of moles of water to the anhydrous salt is= 0.296 / 0.01646 =17.989 ~ 18 So x =18