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)
Please help me. I tried everything. =(
*Convert 1.750g of Fe(NH4)2(SO4)2•6H2O to moles by dividing by the molar mass of this compound.
*Find the moles of H2C2O4 using:
moles = (liters)(mol/L)
Te moles ratio of the two reagents id 1 to 1. That means that the smaller number of moles belongs to the limiting reagent.
To get the theoretical yield of FeC2O42H2O, multiply the moles of the limiting reagent by the molar mass of this product.
To find the theoretical yield of FeC2O42H2O, you need to calculate the number of moles of Fe(NH4)2(SO4)2 6H2O and H2C2O4, then determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, limiting the amount of product that can be formed.
1. Calculate the number of moles of Fe(NH4)2(SO4)2 6H2O:
Given mass = 1.750 g
Molar mass of Fe(NH4)2(SO4)2 6H2O = (55.85 g/mol + 28.04 g/mol + 32.07 g/mol + 4*(1.01 g/mol) + 4*(16.00g/mol)) + 6*(2.02 g/mol + 16.00 g/mol)
Molar mass of Fe(NH4)2(SO4)2 6H2O = 392.14 g/mol
Number of moles of Fe(NH4)2(SO4)2 6H2O = Mass / Molar mass
Number of moles of Fe(NH4)2(SO4)2 6H2O = 1.750 g / 392.14 g/mol
2. Calculate the number of moles of H2C2O4:
Given volume = 13 mL = 0.013 L
Molarity of H2C2O4 = 1.0 M
Number of moles of H2C2O4 = Molarity x Volume
Number of moles of H2C2O4 = 1.0 M x 0.013 L
3. Determine the limiting reactant:
Compare the mole ratio of Fe(NH4)2(SO4)2 6H2O to H2C2O4 using the balanced equation:
1 mole of Fe(NH4)2(SO4)2 6H2O reacts with 1 mole of H2C2O4
The reactant that has the fewer moles compared to the ratio is the limiting reactant.
4. Calculate the theoretical yield of FeC2O42H2O:
From the balanced equation:
1 mole of Fe(NH4)2(SO4)2 6H2O produces 1 mole of FeC2O42H2O
Molar mass of FeC2O42H2O = (55.85 g/mol) + 2*(12.01 g/mol) + 4*(16.00 g/mol) + 2*(1.01 g/mol) + 4*(1.01 g/mol)
Molar mass of FeC2O42H2O = 179.88 g/mol
The theoretical yield can be calculated using the limiting reactant:
Theoretical yield of FeC2O42H2O = Number of moles of limiting reactant x Molar mass of FeC2O42H2O
Remember to convert moles to grams if the units of the molar mass are in grams.
I hope this helps you find the answer to your question.