Solveeit Logo

Question

Question: How to find the theoretical yield for two solid reactants?...

How to find the theoretical yield for two solid reactants?

Explanation

Solution

The balanced chemical equation is used to determine the mole ratio between the reactants and products.From the mole ratio using the molar masses of the species, the theoretical yield can be determined. The theoretical yield is nothing but the mass of the pure sample.

Complete solution:
Consider the reaction ammonia formation as follows:
N2(g) + H2(g)NH3(g){{\text{N}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right) \to \,{\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)
The balanced chemical reaction is as follows:
N2(g) + 3H2(g)2NH3(g){{\text{N}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}\,{\text{3}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right) \to \,2{\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)
The mole ratio between the reactants and the products is as follows:
N2:H2:NH3{{\text{N}}_{\text{2}}}{\text{:}}\,\,\,\,{{\text{H}}_{\text{2}}}\,\,\,{\text{:}}\,\,\,{\text{N}}{{\text{H}}_{\text{3}}}
1mol:3mol:2mol{\text{1}}\,\,{\text{mol}}\,\,\,{\text{:}}\,\,\,{\text{3}}\,\,{\text{mol}}\,\,\,{\text{:}}\,\,{\text{2mol}}
If it is given that 8 grams of the nitrogen react with hydrogen the mass of the ammonia produced then calculated as follows:
The mole ratio between the nitrogen and the ammonia is 1:2.
1molN2 = 2molNH3{\text{1}}\,\,{\text{mol}}\,\,{{\text{N}}_{\text{2}}}\,{\text{ = }}\,\,{\text{2mol}}\,{\text{N}}{{\text{H}}_{\text{3}}}
The molar mass of the nitrogen molecule is 14g{\text{14}}\,{\text{g}} and ammonia is 17g{\text{17}}\,{\text{g}}.
1molN2 = 2molNH3{\text{1}}\,\,{\text{mol}}\,\,{{\text{N}}_{\text{2}}}\,{\text{ = }}\,\,{\text{2mol}}\,{\text{N}}{{\text{H}}_{\text{3}}}
The moles of the substance is calculated by taking the ratio of the mass and its molar mass.
massofN2molarmassofN2 = 2×massofNH3molarmassofNH3\dfrac{{{\text{mass}}\,{\text{of}}\,\,{{\text{N}}_{\text{2}}}}}{{{\text{molar}}\,{\text{mass}}\,{\text{of}}\,\,{{\text{N}}_{\text{2}}}}}\,{\text{ = }}\,\,{\text{2}} \times \dfrac{{{\text{mass}}\,{\text{of}}\,{\text{N}}{{\text{H}}_{\text{3}}}}}{{{\text{molar}}\,{\text{mass}}\,{\text{of}}\,\,{\text{N}}{{\text{H}}_{\text{3}}}}}
Rearrange the equation for the mass of ammonia and substitute the molar mass values of nitrogen and ammonia.
massofNH3=12×massofN2×molarmassofNH3molarmassofN2{\text{mass}}\,{\text{of}}\,{\text{N}}{{\text{H}}_{\text{3}}} = \dfrac{1}{2} \times \dfrac{{{\text{mass}}\,{\text{of}}\,\,{{\text{N}}_{\text{2}}} \times {\text{molar}}\,{\text{mass}}\,{\text{of}}\,\,{\text{N}}{{\text{H}}_{\text{3}}}}}{{{\text{molar}}\,{\text{mass}}\,{\text{of}}\,\,{{\text{N}}_{\text{2}}}}}
massofNH3=12×8g×17g14{\text{mass}}\,{\text{of}}\,{\text{N}}{{\text{H}}_{\text{3}}} = \dfrac{1}{2} \times \dfrac{{{\text{8}}\,{\text{g}} \times {\text{17}}\,{\text{g}}}}{{{\text{14}}}}
massofNH3 = 4.86g{\text{mass}}\,{\text{of}}\,{\text{N}}{{\text{H}}_{\text{3}}}{\text{ = }}\,{\text{4}}{\text{.86}}\,{\text{g}}
Thus, the theoretical yield of the ammonia produced from the 8 grams of nitrogen is 4.86g{\text{4}}{\text{.86}}\,{\text{g}}.
In this way, the theoretical yield can be determined.
Similarly, the mass of the hydrogen is given in the question by using the mole ratio we can determine the theoretical yield of that reactant also.

Note: Determine the yield means we have to determine the amount of the substance. In a chemical reaction, the yield is determined by using the stoichiometry between the reactants and the products.
Stoichiometry is nothing but the mole ratio of reactants and products involved in the balanced chemical reaction.
Here, the mole ratio is typically the ratio of integer numbers. In many chemical reactions, the stoichiometric relation is used to determine the amount that is the yield of the product.