Question

How much heat is evolved when 907kg of Ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure) ${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g);\Delta H = - 91.8kJ$

Answer 1

This could be simply solved by applying the concepts of mole calculations. Also, we know that limiting reagent in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed, so here we need these concepts. We shall calculate the moles of ammonia produced and use it to calculate the heat released.

Formula used: Here, we will use the number of moles formula:
${\text{no of moles = }}\dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$

Complete step by step answer
We are given that 907Kg of $N{H_3}$ reacts.
Given equation:
${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g);$
We are going to identify the moles of respective reactants.
Moles of $N{H_3}$ = $\dfrac{{907000{\text{g}}}}{{18.039{\text{g}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}}} = 5.03 \times {10^4}{\text{mol}}$
Next, divide $\Delta H$ by 2 to get it in terms of one mole of $N{H_3}$ .
${\text{ - }}\dfrac{{{\text{91}}{\text{.8}}}}{{\text{2}}}{\text{ = 45}}{\text{.9kJ/mol of N}}{{\text{H}}_{\text{3}}}$
Finally, multiply −45.9kJ by the number of moles ammonia.
$- 45.9 \times \left( {5.03 \times {{10}^4}} \right) = - 231 \times {10^6}{\text{kJ}}$
Thus, energy released = $231 \times {10^6}{\text{kJ}}$ .

Note
It should be always kept in mind that limiting reagent is defined as the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed. Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. In much the same way, a reactant in a chemical reaction can limit the amounts of products formed by the reaction. When this happens, we refer to the reactant as the limiting reactant (or limiting reagent).
Also, Enthalpy is a concept used in science and engineering when heat and work need to be calculated. When a substance changes at constant pressure, enthalpy tells how much heat and work was added or removed from the substance. Enthalpy is similar to energy, but not the same.