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satela [25.4K]
3 years ago
13

Determine the amount of heat (in kJ) given off when 1.26 × 104 g of ammonia are produced according to the equation N2(g) + 3H2(g

) ⟶ 2NH3(g) ΔH°rxn = −92.6 kJ/mol Assume that the reaction takes place under standardstate conditions at 25°C.
Chemistry
1 answer:
WARRIOR [948]3 years ago
7 0

<u>Answer:</u> The enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

<u>For ammonia:</u>

Given mass of ammonia = 1.26\times 10^4g=1260g

Molar mass of ammonia = 17 g/mol

Putting values in above equation, we get:

\text{Moles of ammonia}=\frac{1260g}{17g/mol}=74.11mol

We are given:

Moles of ammonia = 74.11 moles

For the given chemical reaction:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g);\Delta H^o_{rxn}=-92.6kJ

By Stoichiometry of the reaction:

If 2 moles of ammonia produces -92.6 kJ of energy.

Then, 74.11 moles of ammonia will produce = \frac{-92.6kJ}{2mol}\times 74.11mol=-3431.3kJ of energy.

Thus, the enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.

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