4.0 gram of ferrous ammonium sulphate, FeSO4*(NH4)2SO4 * 6H2O, is used. Since the oxalate is in excess , Calculate the theoretical yield of the iron complex
If you know the equation, post it and I can help you through it. You don't give enough information about the problem for me to write the equation. "Since the oxalate is in excess" doesn't tell me what the oxalate is or what the iron complex is.
FeSO4•NH4)2SO4•6H2O + H2C2O4•2H20 --> FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O
6FeC2O4 + 3H2O2 + 6K2C2O4•H2O -->
4K3[Fe(C2O4)3]•3H2O + 2Fe(OH)3 + 6H2O
2Fe(OH)3 + 3H2C2O4•2H2O + 3K2C2O4•H2O ---> 2K3[Fe(C2O4)3]•3H2O + 9H2O
those are the equations DrBob222
To calculate the theoretical yield of the iron complex, we first need to determine the molar mass of the complex.
The molar mass of ferrous ammonium sulfate is calculated as follows:
FeSO4*(NH4)2SO4 * 6H2O
FeSO4: 55.845 g/mol
(NH4)2SO4: 132.139 g/mol
6H2O: 108.014 g/mol
Adding up the molar masses gives us:
Molar mass = (55.845 g/mol) + 2 × (14.007 g/mol + 1.007 g/mol) + 32.06 g/mol + 6 × (1.008 g/mol + 16.00 g/mol)
= 55.845 g/mol + 2 × (14.007 g/mol + 1.007 g/mol) + 32.06 g/mol + 6 × (1.008 g/mol + 16.00 g/mol)
= 55.845 g/mol + 2 × 15.014 g/mol + 32.06 g/mol + 6 × 17.008 g/mol
= 55.845 g/mol + 2 × 15.014 g/mol + 32.06 g/mol + 102.048 g/mol
= 55.845 g/mol + 30.028 g/mol + 32.06 g/mol + 102.048 g/mol
= 219.981 g/mol
Next, we can calculate the number of moles of ferrous ammonium sulfate used:
Moles = mass / molar mass
= 4.0 g / 219.981 g/mol
≈ 0.01818 mol
Since the oxalate is in excess, the reaction will go to completion and all of the ferrous ammonium sulfate will react. Therefore, the number of moles of the iron complex formed will be equal to the number of moles of ferrous ammonium sulfate used.
Theoretical yield = number of moles × molar mass of the complex
= 0.01818 mol × 219.981 g/mol
≈ 4.0 g
Therefore, the theoretical yield of the iron complex in this reaction is approximately 4.0 grams.
To calculate the theoretical yield of the iron complex, we need to determine the limiting reagent in the reaction between ferrous ammonium sulphate (FeSO4*(NH4)2SO4 * 6H2O) and the oxalate.
To begin, we need to determine the moles of ferrous ammonium sulphate used. To do this, we will use its molar mass:
FeSO4*(NH4)2SO4 * 6H2O
Fe: 1 atom x 55.84 g/mol = 55.84 g/mol
S: 1 atom x 32.06 g/mol = 32.06 g/mol
O: 4 atoms x 16.00 g/mol = 64.00 g/mol
N: 4 atoms x 14.01 g/mol = 56.04 g/mol
H: 28 atoms x 1.01 g/mol = 28.28 g/mol
Total Molar Mass: 236.22 g/mol
Given that you used 4.0 grams of ferrous ammonium sulphate, we can convert this to moles by dividing the mass by the molar mass:
4.0 g / 236.22 g/mol = 0.0169 mol
Next, we need to determine the moles of oxalate used. Since it is in excess, we assume that all of it reacts, and therefore, it is not the limiting reagent.
Finally, we need to calculate the theoretical yield of the iron complex. The balanced equation for the reaction might be helpful:
FeSO4*(NH4)2SO4 * 6H2O + (NH4)2C2O4 -> FeC2O4 + (NH4)2SO4 + H2O
From the balanced equation, we see that there is a 1:1 stoichiometric ratio between FeSO4*(NH4)2SO4 * 6H2O and FeC2O4. Therefore, the moles of FeC2O4 formed would be the same as the moles of ferrous ammonium sulphate used.
The theoretical yield of the iron complex (FeC2O4) would then be:
0.0169 mol (the moles of ferrous ammonium sulphate used)
To find the mass of the theoretical yield, we multiply the number of moles by the molar mass:
0.0169 mol x 111.08 g/mol (molar mass of FeC2O4) = 1.88 g
Therefore, the theoretical yield of the iron complex (FeC2O4) is 1.88 grams.