Chemistry Stoichiometry solutions

Question

Consider mixing 100. mL of 7.50×10^–3M Fe₂(SO₄)₃(aq) [=0.750 mmol Fe₂(SO₄)₃(aq)]

with 50.0 mL of 6.00×10^–2M Ca(OH)₂(aq) [=3.00 mmol Ca(OH)₂(aq)]

that react according to the net ionic precipitation reaction

Fe³⁺(aq) + 3 OH^–(aq) → Fe(OH)₃(s)

Assuming the final volume after the complete reaction is 150 mL, calculate the final concentration of OH^–(aq),Fe³⁺(aq), OH^–(aq) in molarity.

Answer 1

Complete molecular equation: Fe₂(SO₄)₃(aq) + 3Ca(OH)₂(aq) ==> 3CaSO₄(aq) + 2Fe(OH)₃(s)

NIE: 2Fe³⁺(aq) + 6OH^-(aq) ==> 2Fe(OH)₃(s) = Fe³⁺(aq) + 3OH^-(aq) ==> Fe(OH)₃(s)

Find the limiting reactant. An easy way is to divided moles of each reactant by the coefficient in the balanced net ionic equation.

For Fe³⁺: 100. ml Fe₂(SO₄)₂ x 1 L/1000 ml x 7.50x10^-4 mol/L x 2 mol Fe³⁺/mol = 1.50x10^-4mols Fe^÷3+

For OH^-: 50.0 ml Ca(OH)₂ x 1 L/1000 ml x 6.00x10^-2 mol/L x 2 mol OH^-/mol = 6.00x10^-3mols OH^-

÷1.50x10^-4mols Fe^÷3+ by 1 = 1.50x10^-4 for Fe^÷3+

÷6.00x10^-2 by 3 = 2x10^-3 = 2.00x10^-3 for OH^-

Since Fe³⁺ is limiting, we will use the moles of Fe³⁺ to calculate the moles of Fe(OH)₃ produced:

7.5x10^-4 mol Fe³⁺ x 1 mol Fe(OH)₃ / mol Fe³⁺ = 7.50x10^-4 mols Fe(OH)₃

Finally, we can now calculate the moles and concentration of OH^- left in solution

7.5x10^-4 mol Fe³⁺ x 3 mol OH^- / mol Fe³⁺ = 2.25x10^-3 mols OH^- used up

6.00x10^-3 mol OH^- - 2.25x10^-3 mols used = 3.75x10^-3 mol OH^- left over

Final concentration (M) of OH^- = 3.75x10^-3 moles / 150. ml x 1000 ml / L = 0.025 moles / L = 0.025 molar OH^-

Final concentration of Fe³⁺ = 0 molar (it is all in the Fe(OH)₃ precipitate

(be sure to check the math. I did this hurriedly)