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3 years ago

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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.

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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