Chemistry (Theoretical Value) Problem

Question

HNO₃ and Fe₂S₃ react according to the following unbalanced chemical equation:

HNO₂ + Fe₂S₂ → Fe(NO₃)₃ + H₂S

Calculate the theoretical yield (in grams) of Fe(NO₃)₃ in the reaction between 0.711 g HNO₃ and 0.347 g Fe₂S₃.

Answer 1

Hello Daisy

 

Let's first write the balanced chemical reaction.

 

6HNO₃ + Fe₂S₃ → 2Fe(NO₃)₃ + 3H₂S

 

To determine the theoretical yield of Fe(NO₃)₃, we first need to determine which reactant is the limiting reagent using stoichiometry.  Convert the mass of HNO₃ and Fe₂S₃ to moles of Fe(NO₃)₃.  The reactant that produces the lowest moles of Fe(NO₃)₃ is the limiting reagent.

 

Molar mass of HNO₃ = 63 g/mol, Fe₂S₃ = 208 g/mol, Fe(NO₃)₃ = 242 g/mol.  Mole ratio of HNO₃ to Fe(NO₃)₃ is 6:2 and for Fe₂S₃ to Fe(NO₃)₃, it is 1:2.

 

0.711 g HNO₃ x (1 mol HNO₃ / 63 g HNO₃) x (2 mol Fe(NO₃)₃ / 6 mol HNO₃) = 0.00752 mol Fe(NO₃)₃

 

0.347 g Fe₂S₃ x (1 mol Fe₂S₃ / 208 g Fe₂S₃) x (2 mol Fe(NO₃)₃ / 1 mol Fe₂S₃) = 0.00334 mol Fe(NO₃)₃

 

From above, you can see that Fe₂S₃ produces the lower amount of Fe(NO₃)₃, therefore, it is the limiting reagent.

 

Theoretical yield = 0.00334 mol Fe(NO₃)₃ x (242 g Fe(NO₃)₃ / 1 mol Fe(NO₃)₃) = 0.808 g Fe(NO₃)₃