Question

What is the percent yield of the solid product when 12.01 g of iron(III) nitrate reacts in solution with excess sodium phosphate and 5.008 g of the precipitate is experimentally obtained?

Fe(NO₃)₃(aq) + Na₃PO₄(aq) --> FePO₄(s) + NaNO₃(aq) [unbalanced]

Answer 1

A balanced chemical equation is,

Fe(NO₃)₃ (aq) + Na₃PO₄ (aq) -----------> FePO₄ (s) + 3 NaNO₃ (aq)

Given that Na₃PO₄ (aq is used excess and hence Fe(NO₃)₃ (aq) is a limiting reagent and hence reaction yield is governed by Fe(NO₃)₃ (aq) alone.

From reaction stoichiometry it's clear that

1 mole of Fe(NO₃)₃ (aq) 1 mole of FePO₄ (s)

Molar mass of Fe(NO₃)₃ = 241.86 g/mole

Molar mass of FePO₄ (s)  = 150.81 g/mole

Hence we write mass relation as,

241.86 g of Fe(NO₃)₃ (aq) 150.81 g of  FePO₄ (s)

Mass of 1 mole of Fe(NO₃)₃ used = 12.01 g

Let us calculate expected yield of the product  FePO₄ (s) using 12.01 g of 1 mole of Fe(NO₃)_3.

If, 241.86 g of Fe(NO₃)₃ (aq) 150.81 g of  FePO₄ (s)

Then, 12.01 g of Fe(NO₃)₃ (aq) say 'A' g of  FePO₄ (s)

On cross multiplication,

A x 241.86 = 12.01 x 150.81

A = 12.01 x 150.81 / 241.86

A = 7.49 g

Expected or Theoretical Yield of FePO₄ (s) is 7.489 g.

But Actual or Experimental Yield of FePO₄ (s) = 5.008 g ......... (given)

% Yield of FePO₄ (s) = (Actual Yield /Theoretical Yield) x 100

= (5.008/7.489) x 100

% Yield of FePO₄ (s) = 66.87 %

Percent Yield of % Yield of FePO₄ (s) is 66.87 %

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