Answer to Question #342306 in Chemistry for Stev

Question

When Iron 2 hydroxide is mixed with phosphoric acid, iron 2 phosphate and water are formed. Write the balance equation foe this reaction. If 3.20g of iron 2 hydroxide I'd treated with 2.50g of phosphoric acid, what is the limiting reagent

Answer 1

Solution:

The molar mass of Fe(OH)₂ is 89.86 g/mol

The molar mass of H₃PO₄ is 97.994 g/mol

Calculate moles of each reactant:

(3.20 g Fe(OH)₂) × (1 mol Fe(OH)₂ / 89.86 g Fe(OH)₂) = 0.0356 mol Fe(OH)₂

(2.50 g H₃PO₄) × (1 mol H₃PO₄ / 97.994 g H₃PO₄) = 0.0255 mol H₃PO₄

Balanced chemical equation:

3Fe(OH)₂ + 2H₃PO₄ → Fe₃(PO₄)₂ + 6H₂O

According to stoichiometry:

3 mol of Fe(OH)₂ react with 2 mol of H₃PO₄

Therefore, 0.0356 mol of Fe(OH)₂ react with:

(0.0356 mol Fe(OH)₂) × (2 mol H₃PO₄ / 3 mol Fe(OH)₂) = 0.0237 mol H₃PO₄

However, initially there is 0.0255 mol of H₃PO₄ (according to the task).

Thus, Fe(OH)₂ acts as limiting reagent and H₃PO₄ is excess reagent.

Answer:

Balance equation for the reaction: 3Fe(OH)₂ + 2H₃PO₄ → Fe₃(PO₄)₂ + 6H₂O

Iron(II) hydroxide (Fe(OH)₂) is the limiting reagent