Question

Industrial production of nitric acid, which is used in many products including fertilizers and explosives, approaches 10 billion kg per year worldwide. The first step in its production is the exothermic oxidation of ammonia, represented by the following equation. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ΔH⁰rxn = −902.0 kJ If this reaction is carried out using 7.056 ✕ 103 g NH3 as the limiting reactant, what is the change in enthalpy?

Answer 1

Answer: $9.361\times 10^{4} kJ$

Explanation:

The balanced chemical equation :

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$  $\Delta H^0_{rxn}=-902.0kJ$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.056\times 10^3g}{17g/mol}=415.1moles$

According to stoichiometry:

4 moles of $NH_3$ produces = 902.0 kJ of energy

415.1 moles of $NH_3$ produces =$\frac{902.0}{4}\times 415.1=9.361\times 10^{4} kJ$ of energy

Thus the change in enthalpy is $9.361\times 10^{4} kJ$