Answer:
47.2 g
Explanation:
Let's consider the following double displacement reaction.
3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl
The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:
44.3 g × (1 mol / 357.48 g) = 0.124 mol
The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:
3 × 0.124 mol = 0.372 mol
The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:
0.372 mol × (126.75 g/mol) = 47.2 g
The quantity of NaOH required to reach the third equivalence point is 20mL.
Using the titration formula,
CaVa/CbVb = Na/Nb
Where,
Ca = concentration of citric acid (0.200 M)
Cb = concentration of NaOH (0.750 M)
Va = Volume of citric acid (25.0 mL)
Vb = volume of NaOH (x mL)
Na = number of reacting mole of citric acid (3)
Nb = number of reacting mole of NaOH (1)
Therefore Vb ( x mL) =CaVaNb/CbNa
= 0.2× 25×3/0.75 ×1
= 15/0.75
Vb ( x mL) = 20 mL
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