Question

The energy required for the chemical reaction of 21.7 grams of nitrogen to produce ammonia is how many kilojoules? If you put something that isn't the answer I will report you.

Answer 1

`Q=-71.1kJ`

Step-by-step explanation:

Hello!

In this case, since the formation of ammonia by starting with nitrogen and therefore hydrogen is:

`N_2+3H_2\rightarrow 2NH_3`

Which has an energy of reaction of:

`\Delta _fH_(NH_3)=-45.90 (kJ)/(molNH_3)`

We can compute the energy required for this reaction by first computing the moles of ammonia yielded by 21.7 grams of nitrogen (28.02 g/mol) via stoichiometry:

`n_(NH_3)=21.7gN_2*(1molN_2)/(28.02gN_2)*(2molNH_3)/(1molN_2)=1.55molNH_3`

Thus, the energy turns out:

`Q=n_(NH_3)\Delta _fH_(NH_3)=1.55molNH_3 * -45.90 (kJ)/(molNH_3)\\\\Q=-71.1kJ`

Best regards!