in the synthesis of FeC2O4 (dot) 2H2O, which reagent would be the limiting reagent? FeSO4(NH4)2SO4 (dot) 6H2O - 2.68g. H2C2O4 (dot) 2H2O - 1.99g.
I just want to know how the H2Os would effect finding the limiting reagent. As in do they change the ratio? The molecular mass?
Thank you!
The H2O molecules (water of crystallization) do not change the ratio (as long as it isn't the ratio of the water molecules); they do change the molar mass of the parent compound. For example, CuSO4 has a molar mass of about 159.6 while CuSO4.5H2O has a molar mass that is 5*18 (about 90) higher for about 249.7 g/mol.
Thank you :)
To determine the limiting reagent in the synthesis of FeC2O4(dot)2H2O, you need to compare the moles of FeSO4(NH4)2SO4(dot)6H2O and H2C2O4(dot)2H2O in the reaction. The balanced equation for the synthesis is:
FeSO4(NH4)2SO4(dot)6H2O + H2C2O4(dot)2H2O -> FeC2O4(dot)2H2O + (NH4)2SO4 + 2H2O
First, let's calculate the moles of each reagent using their given masses and molar masses:
Molar mass of FeSO4(NH4)2SO4(dot)6H2O:
Fe = 55.85 g/mol
S = 32.06 g/mol
N = 14.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass = (55.85 + 2*14.01 + 32.06 + 6*16.00 + 12*1.01) g/mol
Molar mass = 392.14 g/mol
Moles of FeSO4(NH4)2SO4(dot)6H2O = 2.68 g / 392.14 g/mol
Molar mass of H2C2O4(dot)2H2O:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass = (2*12.01 + 2*1.01 + 4*16.00) g/mol
Molar mass = 90.04 g/mol
Moles of H2C2O4(dot)2H2O = 1.99 g / 90.04 g/mol
Now, let's calculate the mole ratio of FeSO4(NH4)2SO4(dot)6H2O to H2C2O4(dot)2H2O in the balanced equation:
Mole ratio = Coefficient of FeSO4(NH4)2SO4(dot)6H2O / Coefficient of H2C2O4(dot)2H2O
Mole ratio = 1 / 1
Mole ratio = 1
Since the mole ratio is 1:1, the number of moles of FeSO4(NH4)2SO4(dot)6H2O and H2C2O4(dot)2H2O are the same.
The H2O molecules in the compounds do not affect the calculation. They do not change the ratio or the molar mass, as they are not involved in the stoichiometry of the reaction. The H2O molecules are present as crystal water, which is released during the synthesis process.
Therefore, both FeSO4(NH4)2SO4(dot)6H2O and H2C2O4(dot)2H2O have an equal number of moles in this reaction, and neither is the limiting reagent.
To determine the limiting reagent in a chemical reaction, you need to compare the number of moles of each reagent and their respective stoichiometric coefficients from the balanced chemical equation.
In the synthesis of FeC2O4(dot)2H2O, the balanced chemical equation is:
FeSO4(NH4)2SO4(dot)6H2O + H2C2O4(dot)2H2O → FeC2O4(dot)2H2O + (NH4)2SO4 + H2SO4 + H2O
From the given information, the mass of FeSO4(NH4)2SO4(dot)6H2O is 2.68g, and the mass of H2C2O4(dot)2H2O is 1.99g.
To determine the number of moles, you need to divide the given mass of each reagent by their respective molar masses.
The molar mass of FeSO4(NH4)2SO4(dot)6H2O:
Fe: 55.845 g/mol
S: 32.06 g/mol
O: 16.00 g/mol (6 atoms)
N: 14.01 g/mol (2 atoms)
H: 1.01 g/mol (12 atoms)
Total molar mass = (55.845 + 32.06 + (16.00 * 6) + (14.01 * 2) + (1.01 * 12)) g/mol ≈ 392.14 g/mol
The molar mass of H2C2O4(dot)2H2O:
H: 1.01 g/mol (4 atoms)
C: 12.01 g/mol (2 atoms)
O: 16.00 g/mol (6 atoms)
Total molar mass = (1.01 * 4) + (12.01 * 2) + (16.00 * 6) g/mol = 126.07 g/mol
Now, calculate the number of moles for each reagent:
Number of moles of FeSO4(NH4)2SO4(dot)6H2O = 2.68g / 392.14 g/mol ≈ 0.0068 mol
Number of moles of H2C2O4(dot)2H2O = 1.99g / 126.07 g/mol ≈ 0.0158 mol
From the balanced chemical equation, the stoichiometric coefficient of FeSO4(NH4)2SO4(dot)6H2O is 1, and the stoichiometric coefficient of H2C2O4(dot)2H2O is also 1 for the formation of FeC2O4(dot)2H2O.
Since the stoichiometric coefficients are the same, the ratio is not affected by the presence of water molecules (H2O) in the chemical formula. Water does not change the molecular mass or the relationship of moles between elements in compounds.
Comparing the number of moles, we can see that FeSO4(NH4)2SO4(dot)6H2O has 0.0068 mol, while H2C2O4(dot)2H2O has 0.0158 mol.
Since FeSO4(NH4)2SO4(dot)6H2O has fewer moles compared to H2C2O4(dot)2H2O, it is the limiting reagent in this reaction.