Asked by Amy

Dr. bob could you please check this work cus i can't seem to get it right

the first step of the synthesis is described by the reaction below. When 1.750 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4, the theoretical yield of FeC2O42H2O is ????
grams.

Fe(NH4)2(SO4)26H2O(s) + H2C2O4(aq)-->
FeC2O42H2O(s) + (NH4)2SO4(aq) + H2SO4(aq) + 4 H2O(l)

1.750g Fe(NH4)2(SO4)2 6H2O x(1 molFe(NH4)2(SO4)2 6H2O/392.1g)(1mol FeC2O42H2O/1mol Fe(NH4)2(SO4)2 6H2O )= 4.46x10^-3 moles of product formed from Fe(NH4)2(SO4)2 6H2O

3 mL(1L/1000ml)(1mol H2C2O4/L)( 1mol FeC2O42H2O/1mol H2C2O4)= 1.3x10^-2moles of product formed from H2C2O4

(4.46x10^-3 moles of FeC2O42H2O)(392.1g FeC2O42H2O/ 1 mol)= 1.75g FeC2O42H2O produced

I GOT THAT WRONG.



Answered by ~christina~
3 mL(1L/1000ml)(1mol H2C2O4/L)( 1mol FeC2O42H2O/1mol H2C2O4)= 1.3x10^-2moles of product formed from H2C2O4
<b>first, that's 13 not 3ml</b>

and looking at this it seems that it's a acid base reaction.
Answered by ~christina~
Um...forget the last sentence about acid base.
Answered by DrBob222
the first step of the synthesis is described by the reaction below. When 1.750 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4, the theoretical yield of FeC2O42H2O is ????
grams.

Fe(NH4)2(SO4)26H2O(s) + H2C2O4(aq)-->
FeC2O42H2O(s) + (NH4)2SO4(aq) + H2SO4(aq) + 4 H2O(l)<b>OK here</b>

1.750g Fe(NH4)2(SO4)2 6H2O x(1 molFe(NH4)2(SO4)2 6H2O/392.1g)(1mol FeC2O42H2O/1mol Fe(NH4)2(SO4)2 6H2O )= 4.46x10^-3 moles of product formed from Fe(NH4)2(SO4)2 6H2O <b>OK here</b>

3 mL(1L/1000ml)(1mol H2C2O4/L)( 1mol FeC2O42H2O/1mol H2C2O4)= 1.3x10^-2moles of product formed from H2C2O4 <b>You typed 3 mL but you calculated 13 and 0.013 mol of FeC2O4.2H2O is correct.</b>

(4.46x10^-3 moles of FeC2O42H2O)(392.1g FeC2O42H2O/ 1 mol)= 1.75g FeC2O42H2O produced <b>Your error is here. You multiplied by the molar mass of Fe(NH4)2(SO4)2.6H2O (of 392.1) but you want to find mass of FeC2O4.2H2O. You should have multilied by the molar mass of FeC2O4.2H2O (179.897) and I get 0.79836 which I would round to 0.798 or 0.7984 depending upon the number of significant figures you are carrying. </b><i>I hope this helps. Thanks for showing your work. It makes it much easier for us to pick out the error if we can see what you did. </i>

I GOT THAT WRONG.
Answered by ~christina~
hm...it's interesting as I had posted the exact same equation she used..but that person had another name.
Answered by DrBob222
Too many students post under different names. They either feel embarrassed to post too many questions under the same name OR they think posting under several names will get an answer faster OR perhaps both of those. It makes things easier for us if the student uses the same name. It makes no difference to us how many questions one posts as long as some effort is put in to solving the problem.
Answered by tarra
hi DrBob..i got confused dealing with this chemical formula.how i want to find the num of moles of 2g of Fe(NH4)2(SO4)2.6H2O??? Can you help me.
Answered by Caro
Hallo Dr.Bob wie sind die Oxidationszahlen von Fe (NH4)(SO4)2
Answered by monika
calculate percentage of water in Na2co3 . H2O
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