Aluminum sulfate exist as the hydrate Al2(SO4)3 all the numbers are subscriptsso

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.