The first step of the synthesis is described by the reaction below. When 2.000 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)
Did you omit the arrow? I assume the first line holds the reactants and the second line holds the products.
Convert 2.000 g of the iron ammonium sulfate hexahydrate to mols. Convert 13 mL of the 1 M oxalic acid to mols. Determine the mols of the product formed FROM EACH of the reactants, TAKE THE SMALLER ONE and that will be the limiting reagent. Usual stoichiometry after that. Post your work if you need further assistance.
the answer is .918
To determine the theoretical yield of FeC2O4·2H2O, we first need to calculate the moles of Fe(NH4)2(SO4)2·6H2O and H2C2O4.
Step 1: Convert mass of Fe(NH4)2(SO4)2·6H2O to moles.
We are given the mass of Fe(NH4)2(SO4)2·6H2O as 2.000 g. To convert this to moles, we need the molar mass of Fe(NH4)2(SO4)2·6H2O.
Molar mass of Fe(NH4)2(SO4)2·6H2O:
- Fe: 1 atom x 55.85 g/mol = 55.85 g/mol
- N: 2 atoms x 14.01 g/mol = 28.02 g/mol
- H: 16 atoms x 1.01 g/mol = 16.16 g/mol
- S: 2 atoms x 32.07 g/mol = 64.14 g/mol
- O: 18 atoms x 16.00 g/mol = 288.00 g/mol
- Total molar mass: 55.85 + 28.02 + 16.16 + 64.14 + 288.00 = 452.17 g/mol
Now, we can calculate the moles of Fe(NH4)2(SO4)2·6H2O:
Moles = Mass / Molar mass
Moles = 2.000 g / 452.17 g/mol
Step 2: Calculate the moles of H2C2O4.
We are given that we have 13 mL (milliliters) of 1.0 M (moles per liter) H2C2O4. To calculate the moles of H2C2O4, we need to use the molarity and volume.
Moles = Molarity x Volume
Moles = 1.0 mol/L x 0.013 L (13 mL = 0.013 L)
Step 3: Determine the limiting reactant.
To find the moles of FeC2O4·2H2O formed, we need to compare the moles of Fe(NH4)2(SO4)2·6H2O and H2C2O4. The reactant that produces fewer moles of the product will be the limiting reactant.
The balanced equation tells us that the stoichiometric ratio between Fe(NH4)2(SO4)2·6H2O and FeC2O4·2H2O is 1:1. Therefore, the moles of FeC2O4·2H2O formed will be the same as the moles of Fe(NH4)2(SO4)2·6H2O.
The moles of FeC2O4·2H2O = moles of Fe(NH4)2(SO4)2·6H2O
Step 4: Calculate the theoretical yield of FeC2O4·2H2O.
Now that we know the moles of FeC2O4·2H2O formed are equal to the moles of Fe(NH4)2(SO4)2·6H2O, we can calculate the mass of FeC2O4·2H2O using the molar mass of FeC2O4·2H2O.
Molar mass of FeC2O4·2H2O:
- Fe: 1 atom x 55.85 g/mol = 55.85 g/mol
- C: 2 atoms x 12.01 g/mol = 24.02 g/mol
- O: 6 atoms x 16.00 g/mol = 96.00 g/mol
- H: 4 atoms x 1.01 g/mol = 04.04 g/mol
- Total molar mass: 55.85 + 24.02 + 96.00 + 4.04 = 179.91 g/mol
Theoretical yield = Moles of FeC2O4·2H2O x Molar mass of FeC2O4·2H2O
Finally, you can substitute the moles of Fe(NH4)2(SO4)2·6H2O from Step 1 into the equation above, and calculate the theoretical yield of FeC2O4·2H2O in grams.