When iron(ii)hydroxide is mixed with phosphoric acid, iron(ii)phosphate precipitate results.
b)if 3.2g of Fe(OH)² is treated with 2.50g of phosphoric acid, what is the limiting reagent and what is the reactant in access?
c)How many grams of Fe³(PO⁴)² precipitate can be formed?
d)if 3.99g of Fe³(PO⁴)² is actually obtained,what is the percentage yield?
To determine the limiting reagent and the reactant in excess, we need to compare the number of moles of each reactant.
a) First, let's calculate the number of moles for each reactant:
moles of Fe(OH)2 = mass (g) / molar mass (g/mol)
moles of Fe(OH)2 = 3.2g / (55.85 + 2(16.00 + 1.01)) g/mol = 0.0289 mol
moles of phosphoric acid = mass (g) / molar mass (g/mol)
moles of phosphoric acid = 2.50g / (3(1.01) + 2(16.00) + 4(16.00)) g/mol = 0.018 mol
b) The limiting reagent is the reactant that will be completely consumed, and the reactant in excess is the one that will remain after the reaction is complete.
Since the stoichiometry of the reaction is 1:2 (1 mol Fe(OH)2 reacts with 2 mol of phosphoric acid), we can see that the reaction requires twice as many moles of phosphoric acid as Fe(OH)2.
Since we have fewer moles of phosphoric acid (0.018 mol) compared to Fe(OH)2 (0.0289 mol), phosphoric acid is the limiting reagent and Fe(OH)2 is the reactant in excess.
c) To determine the maximum amount of Fe3(PO4)2 precipitate that can be formed, we need to use the stoichiometry of the balanced equation. From the reaction:
1 mol of Fe(OH)2 produces 1 mol of Fe3(PO4)2.
moles of Fe3(PO4)2 = moles of limiting reagent (phosphoric acid) = 0.018 mol
mass of Fe3(PO4)2 = moles of Fe3(PO4)2 x molar mass of Fe3(PO4)2
mass of Fe3(PO4)2 = 0.018 mol x (2(55.85) + 3(31.00) + 8(16.00)) g/mol = 4.14 g (rounded to two decimal places)
Therefore, the maximum amount of Fe3(PO4)2 precipitate that can be formed is 4.14 grams.
d) To calculate the percentage yield, we need to compare the actual yield (3.99 g) to the theoretical yield (4.14 g) and calculate the percentage.
percentage yield = (actual yield / theoretical yield) x 100
percentage yield = (3.99 g / 4.14 g) x 100 = 96.14% (rounded to two decimal places)
Therefore, the percentage yield is approximately 96.14%.
To find the limiting reagent and the reactant in excess, we need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction:
Fe(OH)₂ + 2H₃PO₄ → Fe₃(PO₄)₂ + 6H₂O
a) Calculate moles of Fe(OH)₂:
Molar mass of Fe(OH)₂ = 88.85 g/mol + 2 * (1.01 g/mol + 16.00 g/mol) = 88.85 g/mol + 36.02 g/mol = 124.87 g/mol
moles of Fe(OH)₂ = mass / molar mass = 3.2 g / 124.87 g/mol ≈ 0.0256 mol
b) Calculate moles of H₃PO₄:
Molar mass of H₃PO₄ = 1.01 g/mol + 3 * (1.01 g/mol + 16.00 g/mol + 16.00 g/mol + 16.00 g/mol) = 1.01 g/mol + 3 * 64.00 g/mol = 1.01 g/mol + 192.00 g/mol = 193.01 g/mol
moles of H₃PO₄ = mass / molar mass = 2.50 g / 193.01 g/mol ≈ 0.0129 mol
c) Determine the limiting reagent:
Based on the balanced equation, the stoichiometric ratio between Fe(OH)₂ and H₃PO₄ is 1:2. This means that 1 mol of Fe(OH)₂ reacts with 2 mol of H₃PO₄.
Since we have 0.0256 mol of Fe(OH)₂ and 0.0129 mol of H₃PO₄, the molar ratio is not 1:2. Therefore, H₃PO₄ is the limiting reagent.
d) Calculate the amount of reactant in excess:
To find the amount of the excess reactant, we need to determine how much of the limiting reagent got consumed based on the stoichiometry.
From the stoichiometric ratio, we know that every 2 moles of H₃PO₄ react with 1 mole of Fe(OH)₂. Since 0.0129 mol is used, the amount of Fe(OH)₂ required is (0.0129 mol / 2) ≈ 0.00645 mol.
The initial amount of Fe(OH)₂ was 0.0256 mol, so the remaining amount is 0.0256 mol - 0.00645 mol = 0.0192 mol.
e) Calculate the mass of Fe₃(PO₄)₂ formed:
From the balanced equation, we know that the stoichiometric ratio between Fe(OH)₂ and Fe₃(PO₄)₂ is 1:1.
Molar mass of Fe₃(PO₄)₂ = 3 * (55.85 g/mol) + 2 * (31.03 g/mol + 4 * (16.00 g/mol + 16.00 g/mol + 16.00 g/mol)) = 3 * 55.85 g/mol + 2 * (31.03 g/mol + 192.00 g/mol) = 499.25 g/mol
mass of Fe₃(PO₄)₂ = moles * molar mass = 0.0129 mol * 499.25 g/mol ≈ 6.45 g
f) Calculate the percentage yield:
percentage yield = (actual yield / theoretical yield) * 100
actual yield = 3.99 g
theoretical yield (calculated in part e) = 6.45 g
percentage yield = (3.99 g / 6.45 g) * 100 ≈ 61.9%