Question

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(1) When a sample of ferric sulfate hydrate (EeSO4 XH20(s)) is heated to 270 C in an otherwise empty 625-mL container, it decomposes into ferric oxide (EeO(s), M 71.85 g/mol), sulfur trioxide (SO3(g), M = 80.06 g/mol), and water vapor (H20(g) 18.02 g/mol) wwwww = M = FESO4 XH2O(s) FeO(s) SO3(g) x H20 (g) wwwww www When all the FeSO4xH2O(s) in the sample decomposes, 1.80 g of FeO(s) forms and the total pressure of all gases in the 625-mL container at 270 °C is 8.916 atm www (a) [4 points] Calculate the number of moles of SO 3(g) produced. (b) [8 points] Determine the value of x in the formula of the hydrate FeSO4xH20(s) wwwww

Answer 1

SOL. o given - f esor. Azokg) 1s) Fols) it sozlg tall2019) Also to form feols). given when all the reactant dissolve mass of Feols formed = 1.80 gms. moles of Pols) formed = 1.80 2 Molar mass of Feo = 1.80 mole 71-85 from 1 mole of peso4. Atzo s = I mole of Feols) formed and Imole of so₂19) formed also a mole of len 120 lg) formed to Hence, moles of fes 04.532015) = 1.80 mole 71.85 moles of soalg) formed - 1.80 mole 71.85 = 0.025 mole moles of Holg) formed = 20-025x mole 2x1.50 71.15 Also known V = 625 mi : P = 8.916 atm i T= 2704 Appy 44feous kю PV=nRT In 0.025 +0.0250) 0 8.916X 625 0.025(lt) 80.082 x 543 1000 Allt) 25 upon solving

Hence; 2=4 value of a in the formula of the Hydrate F.so4 .2420 = 4 Hence; net reaction, | Feso4. 4460 IS) - Fols) + 5031g) + 4H20 lg
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