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Topic: analysis of do the efforts to calculation the mr of the FeSO4.xH2O and also thus the X value. (Read 36233 times)




You are watching: Fe(nh4)2(so4)2.6h2o

sssssaaaallmaanRegular MemberPosts: 33Mole Snacks: +1/-0

You acquired n(MnO4-) as 6.603384 x 10-3 right? climate multiply this number through 5/3 and also you'll gain the number of moles of FeC2O4.xH2O. However then when you are working the end the molar mass girlfriend get about 18.2 i m sorry will provide you x as around -7.
Aye. I pertained to the same conclusion. Is it feasible that component 1 of my calculations were incorrect? part OneMass Container to add Fe(NH4)2SO4.6H2O 12.824gMass of container much less some Fe(NH4)2SO4.6H2O 3.030gTherefore fixed of Fe(NH4)2SO4.6H2O 9.794gVolume of i graduated flask used 250 cm³Mr the Fe(NH4)2(SO4)2.6H2O 392.16n(Fe2+) = mass / mr therefore: 9.794/392.16 = 0.024974n(KMnO4)=0.024974/5 ( due to the fact that n(KMnO4)= n(Fe2+) x 5)therefore: n(KMnO4)=0.00499c(KMnO4) = nx1000/ 24.80 = 0.2012I supplied a 0.02 rather of 0.2012 in the calculations, and i got a value of x of about 2.167 i m sorry seems around right. What execute you think?
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No, I've done it on mine calculator and also everything checks out. I have actually no idea what is dorn ------EDIT: What is the: Volume of graduated flask offered 250 cm³ for?
No, I've excellent it on mine calculator and everything check out. I have no idea what is dorn ------EDIT: What is the: Volume of i graduated flask offered 250 cm³ for?
Mass Container plus Fe(NH4)2(SO4)2.6H2O 12.824gMass of container less some Fe(NH4)2(SO4)2.6H2O 3.030gTherefore fixed of Fe(NH4)2(SO4)2.6H2O 9.794gVolume of i graduated flask used 250 cm³Mr of Fe(NH4)2(SO4)2.6H2O 392.16I provided this details to work-related out the concentration the Fe(NH4)2(SO4)2.6H2O. And also i presume that because each molecule the the salt includes one ferrous ion, the number of moles of that is precisely the same as number of moles of Fe(NH4)2(SO4)2.6H2O and also so the concentration should also be the exact same right?.


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i just had an Epiphany. Part 1 of my calculations were wrong ns think.It should look favor this:Part OneMass Container to add Fe(NH4)2SO4.6H2O 12.824gMass that container much less some Fe(NH4)2SO4.6H2O 3.030gTherefore mass of Fe(NH4)2SO4.6H2O 9.794gVolume of i graduated flask supplied 250 cm³Mr of Fe(NH4)2(SO4)2.6H2O 392.16n(Fe2+) = massive / grandfather therefore: 9.794/392.16 = 0.024974C(Fe2+) =nx1000/v ( volume is 250 due to the fact that concentration of Fe(NH4)2(SO4)2.6H2O is the exact same as Fe2+) = 0.099896m1 ( conc. Of Fe2+) = 0.099896v1 (Volume the Fe2+) = 25.00v2 ( Volume that MnO4-) = 24.82Using the equation m2=m1v1/5v2 (Using the following equation ...m2=m1v1/5v2Where m1 and also m2 is concentration that Fe(2+) and also MnO4- and also v1 and also v2 is the matching volumes used.) m2= 0.0201240934Part TwoMass of compound offered 0.200 gFinal Bruette reading 32.82 cm³Final Volume the KMnO4 solution offered 32.82 cm³Thus if a sample that W(g)of the steel (II) ethanedioate hydrate needs a volume v ( cm³ ) the MnO4- equipment of concentration m (mol dm-3), climate the variety of moles the MnO4- reacting is mv/1000 and this variety of moles will react with 5mv/3000 moles of FeC2O4.xH2O.Hence W/Mr=5mv/3000So the Mr=600W/mvMoreover, because Mr=143.87 + 18.02xThen x= (Mr-143.87)/18.02m=0.0201240934v=32.80W=0.200Therefore using Mr=600W/mv = 600 x 0.200/ 32.80 x 0.0201240934Mr= 181.7988275And so, utilizing x= (Mr-143.87)/18.02 = (181.7988275-143.87)/18.02= 2.104818398 i beg your pardon is 2.1 If girlfriend spot any mistakes, might you permit me know please?