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..
I do not understand why my answer is not correct. How do you find the answer for this question?
`1. Write the balanced equation for the reaction.
2. calculate the molecular masses of each compound.
3. calculate the moles of the reactant present.
4. Determine from the amounts, and the balanced equation which is the limiting reageant.
5. Using the moles of the limiting reageant, and the mole ratios in the balanced equation, determine the moles of product.
6. Convert the product moles to mass.
To find the theoretical yield of FeC2O42H2O in this synthesis reaction, follow these steps:
1. Write the balanced equation for the reaction:
Fe(NH4)2(SO4)2 · 6H2O + H2C2O4 → FeC2O4 · 2H2O + (NH4)2SO4 + 6H2O
2. Calculate the molecular masses of each compound:
Fe(NH4)2(SO4)2 · 6H2O: (55.85 + 2*(14.01) + 4*(1.01) + 2*(32.07) + 6*(1.01)) = 392.14 g/mol
H2C2O4: (2*(1.01) + 2*(12.01) + 4*(16.00)) = 90.03 g/mol
FeC2O4 · 2H2O: (55.85 + 2*(12.01) + 4*(16.00) + 2*(1.01)) = 179.91 g/mol
3. Calculate the moles of the reactant present:
Moles of Fe(NH4)2(SO4)2 · 6H2O = (mass of Fe(NH4)2(SO4)2 · 6H2O) / (molecular mass of Fe(NH4)2(SO4)2 · 6H2O)
= 1.750 g / 392.14 g/mol
≈ 0.00446 mol
Moles of H2C2O4 = (concentration of H2C2O4) x (volume of H2C2O4)
= (1.0 M) x (0.013 L)
= 0.013 mol
4. Determine the limiting reagent:
Compare the moles of Fe(NH4)2(SO4)2 · 6H2O and H2C2O4.
Since the moles of H2C2O4 (0.013 mol) are less than the moles of Fe(NH4)2(SO4)2 · 6H2O (0.00446 mol), H2C2O4 is the limiting reagent.
5. Using the moles of the limiting reagent and the mole ratios in the balanced equation, determine the moles of product:
From the balanced equation, the mole ratio of Fe(NH4)2(SO4)2 · 6H2O to FeC2O4 · 2H2O is 1:1.
Therefore, the moles of FeC2O4 · 2H2O = moles of H2C2O4 = 0.013 mol.
6. Convert the product moles to mass:
Mass of FeC2O4 · 2H2O = (moles of FeC2O4 · 2H2O) x (molecular mass of FeC2O4 · 2H2O)
= 0.013 mol x 179.91 g/mol
≈ 2.339 g
Therefore, the theoretical yield of FeC2O42H2O is approximately 2.339 grams.
To find the answer, you should follow the steps you mentioned. Let's go through each step in detail:
1. Write the balanced equation for the reaction:
Fe(NH4)2(SO4)2 · 6H2O + H2C2O4 → FeC2O4 · 2H2O + (NH4)2SO4
2. Calculate the molecular masses of each compound:
- Fe(NH4)2(SO4)2 · 6H2O: 392.14 g/mol
- H2C2O4: 90.03 g/mol
- FeC2O4 · 2H2O: 179.91 g/mol
- (NH4)2SO4: 132.14 g/mol
3. Calculate the moles of the reactant present:
- Fe(NH4)2(SO4)2 · 6H2O: 1.750 g / 392.14 g/mol = 0.00447 mol
- H2C2O4: 1.0 M * 13 mL = 0.013 mol (since the concentration is given in moles per liter, and you have 13 mL, you convert to liters by dividing by 1000)
4. Determine the limiting reagent:
To determine the limiting reagent, compare the moles of each reactant with the coefficients in the balanced equation.
From the balanced equation, the ratio of Fe(NH4)2(SO4)2 · 6H2O to FeC2O4 · 2H2O is 1:1.
The molar ratio of H2C2O4 to FeC2O4 · 2H2O is 1:1.
Comparing the moles, we can see that H2C2O4 is the limiting reagent because it has fewer moles (0.013 mol).
5. Using the moles of the limiting reagent and the mole ratios in the balanced equation, determine the moles of product:
Since the balanced equation tells us that the mole ratio of H2C2O4 to FeC2O4 · 2H2O is 1:1, we can say that the moles of FeC2O4 · 2H2O formed will be 0.013 mol.
6. Convert the product moles to mass:
Using the molecular mass of FeC2O4 · 2H2O (179.91 g/mol) and the moles of FeC2O4 · 2H2O (0.013 mol), we can calculate the mass of FeC2O4 · 2H2O formed:
Mass of FeC2O4 · 2H2O = 0.013 mol * 179.91 g/mol = 2.339 g
Therefore, the theoretical yield of FeC2O4 · 2H2O is 2.339 grams.