The first step of the synthesis is described by the reaction below. When 1.000 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)
Since both reactants are given you know that this is a limiting reagent problem. Convert both reactants to moles.
moles Fe(NH4)2(SO4)2.6H2O = 1g/molar mass=??.
moles H2C2O4 = L x M = ??
These molecules reactant in a 1:1 ratio; therefore, the smaller number of moles is the limiting reagent and that is the Fe(NH4)2(SO4).6H2O.
Convert moles Fe(NH4)2(SO4).6H2O to moles of the product using the coefficients in the balanced equation.
Now convert moles product to grams product by moles x molar mass. This is the theoretical yield.
To find the theoretical yield of FeC2O42H2O, we need to determine the limiting reactant in the reaction.
First, let's calculate the moles of Fe(NH4)2(SO4)2 6H2O:
Molar mass of Fe(NH4)2(SO4)2 6H2O = (1 x 55.85) + (2 x (14 + 4 + 32)) + (4 x (32 + 16 + 16 + 16)) + (12 x 1) + (6 x (2 + 1))
= 392.13 g/mol
Moles of Fe(NH4)2(SO4)2 6H2O = mass / molar mass
= 1.000 g / 392.13 g/mol
= 0.00255 mol
Next, let's calculate the moles of H2C2O4:
Molarity of H2C2O4 = 1.0 M
Moles of H2C2O4 = molarity x volume
= 1.0 mol/L x 0.013 L
= 0.013 mol
Based on the balanced equation, the stoichiometric ratio between Fe(NH4)2(SO4)2 6H2O and H2C2O4 is 1:1. Therefore, H2C2O4 is the limiting reactant because it is in a lower stoichiometric quantity than Fe(NH4)2(SO4)2 6H2O.
Now, let's calculate the theoretical yield of FeC2O42H2O:
From the balanced equation, we can see that 1 mol of Fe(NH4)2(SO4)2 6H2O produces 1 mol of FeC2O42H2O.
Moles of FeC2O42H2O = moles of limiting reactant
= 0.013 mol
Finally, let's convert the moles of FeC2O42H2O to grams:
Molar mass of FeC2O42H2O = (55.85) + (2 x 12.01) + (4 x 16.00) + (2 x (1 x 2 + 16))
= 335.88 g/mol
Theoretical yield of FeC2O42H2O = moles x molar mass
= 0.013 mol x 335.88 g/mol
= 4.36 g
Therefore, the theoretical yield of FeC2O42H2O is 4.36 grams.