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.
3 mL(1L/1000ml)(1mol H2C2O4/L)( 1mol FeC2O42H2O/1mol H2C2O4)= 1.3x10^-2moles of product formed from H2C2O4
first, that's 13 not 3ml
and looking at this it seems that it's a acid base reaction.
Um...forget the last sentence about acid base.
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)OK here
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 OK here
3 mL(1L/1000ml)(1mol H2C2O4/L)( 1mol FeC2O42H2O/1mol H2C2O4)= 1.3x10^-2moles of product formed from H2C2O4 You typed 3 mL but you calculated 13 and 0.013 mol of FeC2O4.2H2O is correct.
(4.46x10^-3 moles of FeC2O42H2O)(392.1g FeC2O42H2O/ 1 mol)= 1.75g FeC2O42H2O produced 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. 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 GOT THAT WRONG.
hm...it's interesting as I had posted the exact same equation she used..but that person had another name.
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.
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.
Hallo Dr.Bob wie sind die Oxidationszahlen von Fe (NH4)(SO4)2
calculate percentage of water in Na2co3 . H2O
To check the work, let's go through the calculations step by step:
1. We start with 1.750g of Fe(NH4)2(SO4)2 6H2O. First, we need to convert this mass to moles using the molar mass of Fe(NH4)2(SO4)2 6H2O.
Calculation: 1.750g Fe(NH4)2(SO4)2 6H2O x (1 mol Fe(NH4)2(SO4)2 6H2O / 392.1g Fe(NH4)2(SO4)2 6H2O)
Result: 0.00446 moles of Fe(NH4)2(SO4)2 6H2O
2. Next, we need to determine the moles of FeC2O42H2O produced from the moles of Fe(NH4)2(SO4)2 6H2O. This is done by using the balanced equation.
Calculation: 0.00446 moles Fe(NH4)2(SO4)2 6H2O x (1 mol FeC2O42H2O / 1 mol Fe(NH4)2(SO4)2 6H2O)
Result: 0.00446 moles of FeC2O42H2O from Fe(NH4)2(SO4)2 6H2O
3. Now, let's calculate the moles of FeC2O42H2O produced from the H2C2O4. We are given 13 mL (0.013 L) of 1.0 M H2C2O4. We can use the molarity and volume to find the moles of H2C2O4, and then use the balanced equation to convert to moles of FeC2O42H2O.
Calculation: 0.013 L H2C2O4 x (1 mol H2C2O4 / 1 L) x (1 mol FeC2O42H2O / 1 mol H2C2O4)
Result: 0.013 moles of FeC2O42H2O from H2C2O4
4. Finally, we can calculate the theoretical yield of FeC2O42H2O by multiplying the moles of FeC2O42H2O from both reactants by the molar mass of FeC2O42H2O.
Calculation: (0.00446 moles + 0.013 moles) x (392.1g FeC2O42H2O / 1 mol FeC2O42H2O)
Result: 5.25g FeC2O42H2O
Based on the calculations above, the theoretical yield of FeC2O42H2O is 5.25 grams, not 1.75 grams.