Question

i answered this question incorrectly, can someone please explain what I did wrong? thank you

6. Use the equation to answer the following questions (equation 2) Equation 2: + Fe(OH)3 MM 106.37m → H2SO4 MM-9.08g/m del Fe () M H2O M a. If you react 2.530g of Fe(OH)with 3.32ml of H 50., what is the limiting reagent assuming your desired product is Fea(SO) (show all work)? - (Don't worry about the sig figs, just make sure all of your final numerical answers have three decimal places, .000) 2.530 9 FeCOH)e : Imol refoto) Imol Feel50)3 = 0.024 mol 106.8 750 dolls Fest sous 3.32 ml H₂SO4 : 332 mc x 1.84 g/ml 0.062 moles 98.08 g/m the limiting reagent is b. Calculate the theoretical mass of Fe (50). 2 moles x molar mass 0.011835 x 30403 g/n - 4,0729

Answer 1

Molar mass of Ferric Sulfate is given wrong.

- 2 F low), + 312504 - Folsoyz_+ 6H20 Tay No of males of Fe(OH)3 = Mass = 2.50g Molari May 106.87 glmat = 0.0237 mal ~ 0.024 mal No of holy of H₂SO4= Mass - Volume & Demity Molar Mass Malwr Man = 3.32 ml x 1.84 g/mL 98.08 glmal 0.0623 mal 0.062 mol - In the Balanced xxn I mole of Ferohls sleguires 3 mole Hesoy to react completely is 0.024 mole of Felohh should require 3 0.024 = 0.036 mole Hasoy But no. of males of Hy soy 2 available are here to Hesoy is excess reagent 4 Fecords is limiting Reagente (6) 2 mole Fe con gives 1 male Fea (souls - .. No of males of Feelsou); formed = 0.024 mol • Theoretical mass = 0.012 mal map of Feg (504); = No of modex Melur = 0.012 ml x 399.88 g/mol = 4.798 g Feg (souls is 399.88 g/mal) (Molar mass of