Answer: The enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For ammonia:
Given mass of ammonia = [tex]1.26\times 10^4g=1260g[/tex]
Molar mass of ammonia = 17 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ammonia}=\frac{1260g}{17g/mol}=74.11mol[/tex]
We are given:
Moles of ammonia = 74.11 moles
For the given chemical reaction:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g);\Delta H^o_{rxn}=-92.6kJ[/tex]
By Stoichiometry of the reaction:
If 2 moles of ammonia produces -92.6 kJ of energy.
Then, 74.11 moles of ammonia will produce = [tex]\frac{-92.6kJ}{2mol}\times 74.11mol=-3431.3kJ[/tex] of energy.
Thus, the enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.
Answer:
-34317.56 Kj
Explanation:
=1.26 x 10^4/17
= 741.2 moles
If 2 moles of ammonia gives - 92.6 Kj/mol
What about 741.2 moles
741.2/2 x - 92.6
= - 34317.56 KJ
