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555gag
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Chemical calculation

Post by 555gag »

http://dx.doi.org/10.1016/j.apsusc.2015.02.181
Hello, can i have a question, please. Article above part Experiment, I have problem to calculate amount of TiOSO4 anf Fe2(SO4)3.xH2O. I dont understatnd, how did they mean this in table 1.

i started with TiOSO4
w(tioSO4) = m(TiOSO4)/(m(tioso4)+m (h2o)
m(tiOso4) = w(tiOSO4)*m(H2O) /(1-w(TiOSO4) = 0,807*100/(1-0,807) = 418,13 g

So, a need to dissolve 418,13 g TiOSO4 in 100ml H2O with 10 ml H2SO4
Is that alright, please ?

Thank you for you explanation
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ChenBeier
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Post by ChenBeier »

I dont have access to the link document.

Your math looks ok. But I cannot confirm your numbers. I don't believe you can dissolve 418 g titanoylsulfate in only 100 ml water.

Maybe you can post the table from the document here.
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Post by 555gag »

In a typical process, the defined amount of TiOSO4 (table 1) was dissolved in 100 ml of hot distilled water acidified with 10 ml 98% H2SO4. The pelluid liquid was diluted in 4 L of distilled water, a definied amount of iron(III) sulphate hydrate Fe2(SO4)3*xH2O was added and the solution was mixed with 300 g of urea (see Table 1).

Table 1
Ti:Fe mol.ratio Fe3+ (wt%) Ti4+ (wt%) SO4 2- (wt %)
1:0,125 12,0 80,7 7,3

Thank you so much[/b]
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Post by 555gag »

Ti:Fe = 1:0,125
Fe3+ (wt%) = 12
Ti4+ (wt%)=80,7
SO4 2-(wt%)=7,3
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Post by ChenBeier »

This shows the mol ratio of Titanium and iron and calculated tot the weight percent.

If you want to have this ratio let say for 100 mg then you have to mix 12 mg Fe3+ and 80.7 mg Ti4+ the sulfate you get automatically.

So you calculation has to be how much mg Fe2( SO4)3 correspond to Fe3+ and TiOSO4 to Ti4+.
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Post by 555gag »

Ok, when i got it right.

12mg............M(Fe3+) = 55,845 g/mol
x...................M(Fe2(SO4)3) = 399,88 g/mol
x= 0,085mg Fe2(SO4)3

80,7 mg..............M(Ti4+) = 47,867 g/mol
y........................M(TiOSO4) = 159,94 g/mol
y= 0,29 mg.

So, in 100ml H2o i have to dissolve 0,29 mg TiOSO4 ?

but, if i want this for example 100 g (, it wil be 12 g Fe 3+ and 80,7 g Ti4+

same calculation
x(Fe2(SO4)3) = 85,926 g
y(TiOSO4) = 290 g

So in 100 ml i have to dissolve 290 g TiOSO4 ?
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Post by ChenBeier »

First you have to check you calculation again. If you have 12 mg iron it cannot be less if you have ironsulfate,

1 mol Fe3+ correspond to 0,5 mol of the sulfate.

And you have to figure out what are the soloubilties of these salts. I don't think 290 g can be dissolved. I think there is a limitation what max. is possible.
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Post by 555gag »

Please, can you tell me a solution, i still counting, but it still less.
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Post by 555gag »

the isrt is x = 85,926 mg
y= 260, 29 mg
Isthat right ?
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Post by ChenBeier »

12 mg Fe3+ is how many moles.

This number correspond to the half of it.

How many mg iron sulfate is it then.
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Post by 555gag »

n=mFe3+/MFe3+ = 12/55,845=0,214 mol
so ,m Fe2(SO4)3 = 0,214*2*399,88=171,85 mg
is theat right
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Post by ChenBeier »

It is not the double it is the half. You have to multiply with 0.5 and not 2.
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Post by 555gag »

wrong,
n(Fe2(SO4)3 = n(fe)/2 = Mfe/2*Mfe = 0,107
mFe2(SO4)3 = 0,107*399,88 = 42,78 mg
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Post by ChenBeier »

This is correct now.
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