Question

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.

Answer 1

-34317.56 Kj

Step-by-step explanation:

• Moles of ammonia = mass/molar mass

=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

Answer 2

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

Step-by-step explanation:

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

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

For ammonia:

Given mass of ammonia =
`1.26* 10^4g=1260g`

Molar mass of ammonia = 17 g/mol

Putting values in above equation, we get:

`\text{Moles of ammonia}=(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 =
`(-92.6kJ)/(2mol)* 74.11mol=-3431.3kJ` of energy.

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